Topics You Need To Understand For This Page
basics derivatives 
chain rule 
Although not required to understand implicit differentiation, several practice problems use the derivative that include trigonometric, exponential and logarithmic terms. You can easily skip the problems that contain derivatives of terms that you have not studied yet and come back to them later. 
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Implicit Differentiation 
on this page:
► What is Implicit Differentiation? ► How To Do Implicit Differentiation ► Examples ► Notation

Implicit differentiation is a technique to calculate the derivative when it is not possible to solve an equation explicitly for y. The technique is just a way to apply the chain rule. So, if you know the chain rule, you will pick up this technique right away. On this page, we give you techniques to solve implicit differentiation problems, starting with some basic examples and building in difficulty as the discussion progresses. 
What is Implicit Differentiation? 
Implicit differentiation is a technique to calculate the derivative when there is no way to solve the equation for y. For example, up until now, you probably were given or were able to write an equation like
\( y = \sin(2x) + x^4 + e^{x} \)
Notice that all the variables on the right side of the equal sign are x's and the variable y is alone on the left and has a power of 1. In essence, the equation is solved for a single y. Then, when you take the derivative, you really only need to work with the right side and the answer just falls out as \( dy/dx \).
Sometimes it is just not possible to isolate an individual y. For example, in the equation
\( y + e^{y} = \sin(2x) + x^4 + e^{x} \)
there is no way to solve for y because of the \(e^{y}\) term. In cases like this we use implicit differentiation to calculate \( dy/dx \).
Before we look at exactly how to do implicit differentation, let's look at a few points that will give you an overview of what to look for.
1. We assume that the equation we are given has one independent variable, usually x or t, and the dependent variable, usually y, i.e. y is a function of x or t.
2. The derivative we are asked to determine is \( dy/dx \) or \( dy/dt \).
3. Since the derivative does not automatically fall out at the end, we usually have extra steps where we need to solve for it.
4. The chain rule is used extensively and is a required technique.
5. Implicit differentiation expands your idea of derivatives by requiring you to take the derivative of both sides of an equation, not just one side.
How To Do Implicit Differentiation 
Okay, lets watch a couple of video clips that will explain this concept. This first video is very good. He starts off with a review of the chain rule and then shows how to use the chain rule to perform implicit differentiation.
 MIP4U  Introduction to Basic Implicit Differentiation [6min47secs] 

That first video should be enough to get you started. But if you need more information, here is a video clip that might help.
 PatrickJMT  Implicit Differentiation  Basic Idea and Examples [2min52secs] 

All of this explanation may be a bit theoretical at this point and you may be asking yourself how to use these ideas. So let's do an example before we go on. We start with an example that you already know how to do and build from there.
 Find \(dy/dx\) of \(y=x^2+5\) 
You already know that \( y' = 2x \). That's easy. But we are going to write it a bit differently so that you can see what is REALLY going on and be able to apply the same idea to an implicit differentiation problem.
What is actually happening here is that we are taking the derivative of both sides and applying the chain rule as follows
\(\displaystyle{ \frac{d}{dx}[y] = \frac{d}{dx}[x^2 + 5] }\)
On the left of the equal sign, we have \(\displaystyle{ \frac{d}{dx}[y] }\). Since the y does not match the x in the \(dx\) of the derivative, we can't do anything with it. We apply the chain rule and use the shorthand notation by writing \( dy/dx \).
The right side is easy. First, we use the addition and subtraction rules to write
\(\displaystyle{ \frac{d}{dx}[x^2 + 5] = \frac{d}{dx}[x^2] + \frac{d}{dx}[5] }\)
Since the x in the \( x^2 \) term matches the x in the \(dx\) of the derivative, we can directly use the power rule and get \( 2x \).
This probably seems like a strange way of thinking about it but that is really what is going on. You have been doing shortcuts since you first learned derivatives and may not know it.
Finally, we use the constant rule to get \(\displaystyle{ \frac{d}{dx}[5] = 0 }\).
So we end up with \( dy/dx = 2x \), just as you would expect.
Now, let's do something a bit strange here. Let's rework this same example a little differently so that you can see where implicit differentiation comes in.
Let's rewrite \( y = x^2 + 5 \) as \( y  x^2 = 5 \) and calculate \( dy/dx \) again. As we go, let's apply each of the implicit differentiation ideas 15 that we discussed above.
1. We assume that y is a function of x. This was obvious when we wrote it as \( y = x^2 + 5 \) but not so obvious when we wrote it as \( y  x^2 = 5 \). So it is required that we mention it.
2. We are asked to determine \( dy/dx \) from the problem statement. Again, we need to understand that before we go on.
3. The extra step will show up at the end. We will mention this again when we get there.
4. We use the chain rule when we take the derivative of y with respect to x.
5. Let's start with this idea and work this example again, in this new form.
So, we are asked to find \( dy/dx \) of \( y  x^2 = 5 \). We start by taking the derivative of both sides.
\(\displaystyle{ \frac{d}{dx} [yx^2] = \frac{d}{dx}[5] }\)
There is nothing new here. We use a basic rule of algebra that says we can do the same thing to both sides of an equation.
The right side is easy since the constant rule says that the derivative of a constant is just zero.
On the left side we start with the addition and subtraction rules to get
\(\displaystyle{ \frac{d}{dx} [yx^2] = \frac{d}{dx}[y]  \frac{d}{dx}[x^2] }\)
The second term is easy.
\(\displaystyle{ \frac{d}{dx}[x^2] = 2x }\)
The first term is the one we will focus on. You've seen it before but let's discuss it in detail. Since the y does not match the x in the derivative term \(dx\), we have to use the chain rule. We cannot simplify this so we just write it in shorthand notation as \(dy/dx\). That's it. That's all there is to it. Now, let's lay out all the steps and finish the example.
\(\displaystyle{
\begin{array}{rcl}
\frac{d}{dx} [yx^2] & = & \frac{d}{dx}[5] \\
\frac{d}{dx}[y]  \frac{d}{dx}[x^2] & = & 0 \\
dy/dx 2x & = & 0 \\
dy/dx & = & 2x
\end{array}
}\)
Now we can discuss idea 3 above. Notice that in the second to the last line, we have \(dy/dx  2x = 0\). We are asked in the directions to find \(dy/dx\). The last step above is the extra step we need to do to solve for \(dy/dx\) to get \(dy/dx=2x\).

Okay, before we do another example, let's look at some specific terms you will see. Go through these examples carefully and make sure you understand all the steps here before going on. Most instructors will not go through these details and expect you to pick them up on your own, which, although possible, is not easy. So we are trying to give you a head start by showing these details.
Examples of Individual Terms 
Here are some terms that you may see when you are using implicit differentiation. The techniques used here are not new but these examples show you how to handle the extra yterm.
 Take the derivative of \(y^2\) with respect to \(x\). 
In this example, we will look at how to handle the derivative \(\displaystyle{\frac{d}{dx}[y^2] }\).
Once you understand how to do this, you will be able to handle a yterm whose power is any real number.
Notice, this is very similar to a derivative you already know how to do, i.e. \(\displaystyle{\frac{d}{dx}[x^2] = 2x }\).
The difference is, instead of \(x^2\), we have \(y^2\). Since the y in \(y^2\) does not match the x in \(dx\), we need to use the chain rule.
\(\displaystyle{\frac{d}{dx}[y^2] = 2y \frac{d}{dx}[y] }\).
Now, since we can't do anything with the y in \(\displaystyle{ \frac{d}{dx}[y] }\), we just leave it and carry it along. However, it is easier to write it in shorthand notation as \(\displaystyle{ \frac{dy}{dx} }\) or \(dy/dx\).
That's as far as we can go. So \(\displaystyle{\frac{d}{dx}[y^2] = 2y(dy/dx)}\).

 Use the result from the previous example, to find the derivative of \(e^{y^2}\) with respect to \(x\). 
In this example, we will look at how to handle the derivative \(\displaystyle{\frac{d}{dx}[e^{y^2}] }\)
First, we use the chain rule.
\(\displaystyle{\frac{d}{dx}[e^{y^2}] = e^{y^2} \frac{d}{dx}[y^2] }\)
From the previous example, we know that \(\displaystyle{ \frac{d}{dx}[y^2] = 2y (dy/dx) }\)
So, we have
\(\displaystyle{\frac{d}{dx}[e^{y^2}] = 2y e^{y^2} (dy/dx) }\)
Notice that we applied the chain rule twice.

Okay, so now you are ready to see how we use this technique to determine \(dy/dx\).
 Find \( dy/dx \) of \( y^2  x = 7 \). 
Notice there is no way to solve this equation for y. So implicit differentiation is our only option.
\(\displaystyle{
\begin{array}{rcl}
\frac{d}{dx}[y^2x] & = & \frac{d}{dx}[7] \\
\frac{d}{dx}[y^2]  \frac{d}{dx}[x] & = & 0 \\
2y \frac{d}{dx}[y]  1 & = & 0 \\
2y \frac{dy}{dx}  1 & = & 0 \\
2y \frac{dy}{dx} & = & 1 \\
\frac{dy}{dx} & = & \frac{1}{2y}
\end{array}
}\)
The last two lines implement idea 3 above to solve for \(dy/dx\). This is one of the unique steps that is almost always required in implicit differentiation problems.

Another term you will run across when doing implicit differentiation is \(xy\).
 Take the derivative of \(xy\) with respect to \(x\). 
In this example, we will look at how to handle the derivative \(\displaystyle{\frac{d}{dx}[xy] }\)
First, we use the product rule.
\(\displaystyle{\frac{d}{dx}[xy] = x \frac{d}{dx}[y] + y \frac{d}{dx}[x] }\)
Since we can't simplify \(\displaystyle{ \frac{d}{dx}[y] }\), we write it in shorthand terms as \(dy/dx\).
The last term is easy \(\displaystyle{ \frac{d}{dx}[x] = 1}\).
Finally, we have \(\displaystyle{\frac{d}{dx}[xy] = x (dy/dx) + y }\)

 Use the result from the previous example to find the derivative of \(\sin(xy)\) with respect to \(x\). 
In this example, we will look at how to handle the derivative \(\displaystyle{\frac{d}{dx}[\sin(xy)] }\)
First, we apply the chain rule.
\(\displaystyle{\frac{d}{dx}[\sin(xy)] = \cos(xy) \frac{d}{dx}[xy] }\)
From the previous example, we know that \(\displaystyle{\frac{d}{dx}[xy] = x (dy/dx) + y }\)
So we end up with \(\displaystyle{\frac{d}{dx}[\sin(xy)] = \cos(xy) [x (dy/dx) + y ] }\)

Note   After going through these examples, if you want more complicated examples with lots of explanation, going through the solutions to practice problems will help you.
One last comment . . . some books and instructors use \( y' \) instead of \( dy/dx \) or \( dy/dt \). However, I have found that it is easy to lose the 'prime' mark on \( y' \) in subsequent algebra. I highly recommend that you use \( dy/dx \) instead of \( y' \) as you are learning this technique. The time to use shorthand notation is after you are comfortable with the technique. Of course, you need to follow what your instructor requires.
Instructions   Unless otherwise instructed, calculate dy/dx of the following equations using implicit differentiation. Assume y is a function of x.
 Give your final answers in completely factored form.
 When you are asked to find the equation of tangent lines, give your answers in slopeintercept form.
 When you are asked to graph, use a graphing utility to plot the equation and the tangent line(s) on the same set of axes.
Practice A01 
\(y=e^{xy}\) 


\(\displaystyle{\frac{dy}{dx}=\frac{ye^{xy}}{1xe^{xy}}}\) 
Take the derivative of both sides. 
\(\displaystyle{\frac{d}{dx}[y]}\)  \(=\) 
\(\displaystyle{\frac{d}{dx}\left[ e^{xy} \right]}\) 
Use the chain rule on the exponential term. 
\(\displaystyle{\frac{dy}{dx}}\)  \(=\) 
\(\displaystyle{e^{xy}\frac{d}{dx}[xy]}\) 
Use the product rule. 
 \(=\)  \(\displaystyle{e^{xy} \left[ x\frac{dy}{dx} + y (1) \right]}\) 
 \(=\)  \(\displaystyle{xe^{xy}\frac{dy}{dx} + ye^{xy}}\) 
Move dy/dx terms to the left, other terms to the right. 
\(\displaystyle{\frac{dy}{dx}  xe^{xy}\frac{dy}{dx}}\)  \(=\) 
\(\displaystyle{ye^{xy}}\) 
Factor out dy/dx. 
\(\displaystyle{\frac{dy}{dx} \left[ 1xe^{xy} \right]}\)  \(=\) 
\(\displaystyle{ye^{xy}}\) 
Solve for dy/dx. 
\(\displaystyle{\frac{dy}{dx}}\)  \(=\) 
\(\displaystyle{\frac{ye^{xy}}{1xe^{xy}}}\) 
Practice A01 Final Answer 
\(\displaystyle{\frac{dy}{dx}=\frac{ye^{xy}}{1xe^{xy}}}\) 
Practice A02 
\(x^2+y^2=25\); find \(dy/dx\) and the slope at \(x=3\) 


\(\displaystyle{dy/dx=x/y}\) \(m=3/4\) at \((3,4)\) \(m=3/4\) at \((3,4)\) 
This video shows a quick calculation of the derivative but does not solve the complete problem. The complete, written out solution is shown below the video.
Calculate \(dy/dx\).
\(\displaystyle{
\begin{array}{rcl}
\frac{d}{dx}\left[ x^2 + y^2 \right] & = & \frac{d}{dx}[25] \\
2x + 2y\frac{dy}{dx} & = & 0 \\
2y\frac{dy}{dx} & = & 2x \\
\frac{dy}{dx} & = & x/y
\end{array}
}\)
Find the slope of the tangent line(s).
To find the slope, we are given \(x=3\). We need to find the yvalue.
Using \( x^2 + y^2 = 25 \), we have \( (3)^2 + y^2 = 25 \rightarrow y^2 = 16 \rightarrow y = \pm 4 \). Don't forget the \( \pm \).
So the two possible points are \( (3,4), (3,4) \). This means the slope (m) is \( m = x/y = 3/4 \) or \( m = x/y = 3/4 \). Since we were not given any information about which one to choose (like quadrant or sign), they both need to be included.
Practice A02 Final Answer 
\(\displaystyle{dy/dx=x/y}\) \(m=3/4\) at \((3,4)\) \(m=3/4\) at \((3,4)\) 
Practice A03 
\(2x^3+y^37=0\) 


\(\displaystyle{ dy/dx = 2x^2/y^2 }\) 
\(\displaystyle{
\begin{array}{rcl}
\frac{d}{dx}\left[ 2x^3 + y^3  7 \right] & = & \frac{d}{dx}[0] \\
6x^2 + 3y^2\frac{dy}{dx} & = & 0 \\
3y^2\frac{dy}{dx} & = & 6x^2 \\
\frac{dy}{dx} & = & \frac{6x^2}{3y^2} = \frac{2x^2}{y^2}
\end{array}}\)
Practice A03 Final Answer 
\(\displaystyle{ dy/dx = 2x^2/y^2 }\) 
Practice A04 
For \(xy^2=100\), calculate \(dy/dx\), the tangent line at \((1,10)\) and graph. 


\(\displaystyle{ \frac{dy}{dx} = \frac{y}{2x} }\) The equation of the tangent line at the point \((1,10)\) is \( y=5x+15 \). 
calculate the derivative \( dy/dx \)
take the derivative of both sides 
\(\displaystyle{ \frac{d}{dx}[xy^2] }\) 
\(=\) 
\(\displaystyle{ \frac{d}{dx}[100] }\) 
use the product rule 
\(\displaystyle{ x \frac{d}{dx}[y^2] + y^2 \frac{d}{dx}[x] }\) 
\(=\)  \(0\) 
use the chain rule on the \(d[y^2]/dx \) term 
\(\displaystyle{ x (2y) \frac{dy}{dx} + y^2 (1) }\) 
\(=\)  \(0\) 
separate terms with \(dy/dx\) on the left and terms without \(dy/dx\) to the right of the equal sign 
\(\displaystyle{ (2xy) \frac{dy}{dx} }\)  \(=\) 
\(y^2\) 
divide both sides by \( 2xy \) to solve for \( dy/dx \) 
\(\displaystyle{ \frac{dy}{dx} }\) 
\(=\)  \(\displaystyle{ \frac{y^2}{2xy} }\) 
simplify by canceling \(y\) 
\(\displaystyle{ \frac{dy}{dx} }\) 
\(=\)  \(\displaystyle{ \frac{y}{2x} }\) 
So now we have the derivative \( dy/dx = y/(2x) \) at every point on the curve. [special note about range: Since we divided by \(2xy\) in step 5 and canceled a y in step 6, we need to check that \(y \neq 0 \). We would normally add a special note here about that. However, \(y=0\) is not in the range of the original equation. So canceling y here is okay and nothing additional needs to be said. This is a hidden detail that could trip you up in the future if you are not watching for it.]
 calculate the derivative at the point  \( dy/dx \) at \((1,10)\)
Since the derivative is the slope, this gives us the slope at that point. This step just involves plugging \(x=1\) and \(y=10\).
\(\displaystyle{ \frac{dy}{dx}(1,10) = \frac{10}{2(1)} = 5 }\)
 find the equation of the tangent line 
Since we know the slope \(m=5\) and one point \((1,10)\), we can find the equation of the tangent line. However, we must be careful to check that the point is actually on the curve itself. If it isn't, we need to use a different technique (details here). So, let's check by plugging the point into the original function.
\( xy^2 = 100 \) at \((1,10)\), \( x=1; ~~~ y=10 ~~~ \to ~~~ (1)(10)^2 = 100 ~~~ \to 100 = 100 \)
So this tells us that the point is on the curve, so we can proceed.
We will use the pointslope form of the equation of a line \( yy_1 = m(xx_1) \) where \( (x_1,y_1) = (1,10)\)
\( y10 = (5)(x1) ~~~ \to ~~~ y = 5x+15 \)
 plot the equation of the curve and the tangent line on the same set of axes 
The graph below was produced using winplot, which is a free graphing utility. You can find a link on the Tools And Resources page. The blue line is the tangent line and the red line is the original equation.
Practice A04 Final Answer 
\(\displaystyle{ \frac{dy}{dx} = \frac{y}{2x} }\) The equation of the tangent line at the point \((1,10)\) is \( y=5x+15 \). 
Practice A05 
\(\sin(x/y)=1/2\) 


\(dy/dx = y/x\) 
\(\displaystyle{
\begin{array}{rcl}
\frac{d}{dx}\left[ \sin(x/y) \right] & = & \frac{d}{dx}[ 1/2 ] \\
\cos(x/y) \cdot \frac{d}{dx}[x/y] & = & 0 \\
\cos(x/y) \cdot \frac{y(1)  x(dy/dx)}{y^2} & = & 0 \\
\frac{\cos(x/y)}{y}  \frac{x \cos(x/y)}{y^2}\frac{dy}{dx} & = & 0 \\
\frac{x \cos(x/y)}{y^2}\frac{dy}{dx} & = & \frac{ \cos(x/y)}{y} \\
\frac{dy}{dx} & = & \frac{ \cos(x/y)}{y} \cdot \frac{y^2}{x \cos(x/y)} = y/x
\end{array}}\)
Practice A05 Final Answer 
\(dy/dx = y/x\) 
Practice A06 
\((x+y)^2=xy3\) 


\(\displaystyle{\frac{dy}{dx}=\frac{12x2y}{1+2x+2y}}\) 
Take the derivative of both sides. 
\(\displaystyle{\frac{d}{dx}\left[ (x+y)^2 \right]}\)  \(=\) 
\(\displaystyle{\frac{d}{dx}[ x  y  3 ]}\) 
Use the chain rule. 
\(\displaystyle{2(x+y)\frac{d}{dx}[x+y]}\)  \(=\) 
\(\displaystyle{1  \frac{d[y]}{dx}}\) 
Use implicit differentiation on d[y]/dx. 
\(\displaystyle{2(x+y)(1 + \frac{dy}{dx})}\)  \(=\) 
\(\displaystyle{1  \frac{dy}{dx}}\) 
\(\displaystyle{2(x+y) + 2(x+y)\frac{dy}{dx} + \frac{dy}{dx}}\)  \(=\) 
\(1\) 
Move terms with dy/dx to the left, other terms to the right. 
\(\displaystyle{2(x+y)\frac{dy}{dx} + \frac{dy}{dx}}\)  \(=\) 
\(\displaystyle{1  2(x+y) }\) 
Factor out dy/dx. 
\(\displaystyle{[2(x+y) + 1] \frac{dy}{dx}}\)  \(=\) 
\(\displaystyle{1  2(x+y) }\) 
Solve for dy/dx. 
\(\displaystyle{\frac{dy}{dx}}\)  \(=\) 
\(\displaystyle{\frac{1  2(x+y)}{2(x+y) + 1}}\) 
Practice A06 Final Answer 
\(\displaystyle{\frac{dy}{dx}=\frac{12x2y}{1+2x+2y}}\) 
Practice A07 
For \((xy1)^3=x\), find \(dy/dx\), the tangent line at \((1,1)\) and graph. 


\(\displaystyle{\frac{dy}{dx}=1\frac{1}{3(xy1)^2}}\) the equation of the tangent line is \(y=2x/35/3\) 
calculate the derivative
\(\displaystyle{
\begin{array}{rcl}
\frac{d}{dx}[ (xy1)^3 ] & = & \frac{d}{dx}[x] \\
3(xy1)^2 \frac{d}{dx}[xy1] & = & 1 \\
3(xy1)^2 \left[ 1  \frac{dy}{dx}  0 \right] & = & 1 \\
1  \frac{dy}{dx} & = & \frac{1}{3(xy1)^2} \\
\frac{dy}{dx} & = & 1\frac{1}{3(xy1)^2}
\end{array}}\)
find the equation of the tangent line
Okay, so we found the derivative \(dy/dx\). Now we need to find the equation of the tangent line. If you need a refresher on how to do this, you can find an explanation and examples on the Equations of Tangent Lines page.
First, we need to see if the point \((1,1)\) is on the curve. The result will determine what technique we use. To test this, we substitute the point in the original equation.
\((1,1) ~~ \to ~~ x=1, y=1\)
\((xy1)^3=x ~~ \to ~~ (1(1)1)^3 = 1 ~~ \to ~~ 1^3=1 ~~ \to ~~ 1=1 \)
So, this shows that the point is on the curve. So all we need is the slope. Since the derivative is the slope, we can just substitute the point into the equation we found for \(dy/dx\) as follows.
\(\displaystyle{ \frac{dy}{dx} = 1\frac{1}{3(1+11)^2} = 1\frac{1}{3} = \frac{2}{3}}\)
So now we have the slope \(m=2/3\) and the point \((1,1)\) and we can find the equation. Let's use the pointslope form.
\( y(1) = (2/3)(x1) ~~ \to y = (2/3)x  2/3  1 ~~ \to ~~ y = (2/3)x  5/3 \)
graph
This graph, shown below, was produced using winplot, which you can find on the Tools And Resources page. The black curve is the graph of \((xy1)^3=x\). We used the implicit graphing option to generate this curve.
Practice A07 Final Answer 
\(\displaystyle{\frac{dy}{dx}=1\frac{1}{3(xy1)^2}}\) the equation of the tangent line is \(y=2x/35/3\) 
Practice A08 
For \(e^{xy}+x^2=y^2\), calculate \(dy/dx\), the tangent line at \((0,1)\) and graph. 


\(\displaystyle{\frac{dy}{dx}=\frac{2x+ye^{xy}}{2yxe^{xy}}}\) The equation of the tangent line is \(y=x/2+1\). 
First we need to find \(dy/dx\).
\(\displaystyle{
\begin{array}{rcl}
\frac{d}{dx}[e^{xy}+x^2] & = & \frac{d}{dx}[y^2] \\
e^{xy}\frac{d}{dx}[xy] + 2x & = & 2y \frac{dy}{dx} \\
e^{xy}\left[ x\frac{dy}{dx} +y(1) \right] + 2x & = & 2y\frac{dy}{dx} \\
xe^{xy}\frac{dy}{dx} + ye^{xy} + 2x  2y\frac{dy}{dx} & = & 0 \\
xe^{xy} \frac{dy}{dx}  2y \frac{dy}{dx} & = & 2x  ye^{xy} \\
( xe^{xy}  2y ) \frac{dy}{dx} & = & 2x  ye^{xy} \\
\frac{dy}{dx} & = & \frac{2xye^{xy}}{xe^{xy}2y} \\
& = & \frac{2x+ye^{xy}}{2yxe^{xy}}
\end{array}}\)
Okay, so now that we have \(dy/dx\), we need to find the slope at the point \((0,1)\). The result \(dy/dx\) is the slope at every point on the curve, so to get the slope at the specific point \((0,1)\), we just substitute the point in the expression for \(dy/dx\).
\(\displaystyle{\frac{dy}{dx} = \frac{2(0)+1e^0}{2(1)0e^0} = \frac{1}{2}}\)
Now to find the equation of the tangent line, let's use the slopeintercept form, \(y=mx+b\).
\( 1 = (1/2)(0)+b ~~ \to ~~ b=1\)
So the equation of the tangent line is \( y = x/2+1 \).
This graph was produced using winplot, which you can find on the Tools And Resources page.
The black curve is the graph of \(e^{xy}+x^2=y^2\). We used the implicit graphing option to generate this curve. Notice that this curve is NOT a function since it does not pass the vertical line test.
Practice A08 Final Answer 
\(\displaystyle{\frac{dy}{dx}=\frac{2x+ye^{xy}}{2yxe^{xy}}}\) The equation of the tangent line is \(y=x/2+1\). 
Practice A09 
find tangent line of \(y\sin(2x)=x\cos(2y)\) at \((\pi/2,\pi/4)\). 


the equation of the tangent line is \(y=x/2\) 
Practice A09 Final Answer 
the equation of the tangent line is \(y=x/2\) 
Practice A10 
\(1+x=\sin(xy^2)\) 


\(\displaystyle{\frac{dy}{dx}=\frac{1y^2\cos(xy^2)}{2xy\cos(xy^2)}}\) 
Practice A10 Final Answer 
\(\displaystyle{\frac{dy}{dx}=\frac{1y^2\cos(xy^2)}{2xy\cos(xy^2)}}\) 
Practice A11 
\(\displaystyle{x^2+xy+y^2=3}\); find \(dy/dx\) and the tangent line at \((1,1)\) 


\(\displaystyle{\frac{dy}{dx}=\frac{2xy}{x+2y^2}}\) the equation of the tangent line at \((1,1)\) is \(y=x+2\) 
Practice A11 Final Answer 
\(\displaystyle{\frac{dy}{dx}=\frac{2xy}{x+2y^2}}\) the equation of the tangent line at \((1,1)\) is \(y=x+2\) 
Practice A12 
\(x^3+y^3=xy\) 


\(\displaystyle{\frac{dy}{dx}=\frac{y3x^2}{3y^2x}}\) 
Practice A12 Final Answer 
\(\displaystyle{\frac{dy}{dx}=\frac{y3x^2}{3y^2x}}\) 
Practice A13 
\(x^2y+xy^2=3x\) 


\(\displaystyle{\frac{dy}{dx}=\frac{32xyy^2}{x^2+2xy}}\) 
Practice A13 Final Answer 
\(\displaystyle{\frac{dy}{dx}=\frac{32xyy^2}{x^2+2xy}}\) 
Practice A14 
\(\displaystyle{x^{2}\ln(y^3)=7}\) 


\(\displaystyle{\frac{dy}{dx}=\frac{2y}{3x^3}}\) 
Practice A14 Final Answer 
\(\displaystyle{\frac{dy}{dx}=\frac{2y}{3x^3}}\) 
Practice A15 
find \(ds/dt\) for \(s^3t^3=8\) 


\(\displaystyle{\frac{ds}{dt}=t^2/s^2}\) 
Practice A15 Final Answer 
\(\displaystyle{\frac{ds}{dt}=t^2/s^2}\) 
Practice A16 
find \(dy/dt\) for \(y^3+\sin(yt)=t^5\) 


\(\displaystyle{\frac{dy}{dt}=\frac{5t^4y\cos(yt)}{3y^2+t\cos(yt)}}\) 
Practice A16 Final Answer 
\(\displaystyle{\frac{dy}{dt}=\frac{5t^4y\cos(yt)}{3y^2+t\cos(yt)}}\) 
Practice A17 
\(x\sin(y)+y\sin(x)=1\) 


\(\displaystyle{dy/dx=\frac{\sin(y)y\cos(x)}{x\cos(y)+\sin(x)}}\) 
Both of these videos contain the solution to this problem. We are including both here in case one of them helps you more than the other.
Practice A17 Final Answer 
\(\displaystyle{dy/dx=\frac{\sin(y)y\cos(x)}{x\cos(y)+\sin(x)}}\) 
Practice A18 
\(\displaystyle{2x+3e^y=e^{x+y}}\) 


\(\displaystyle{\frac{dy}{dx}=\frac{e^{x+y}2}{3e^ye^{x+y}}}\) 
Practice A18 Final Answer 
\(\displaystyle{\frac{dy}{dx}=\frac{e^{x+y}2}{3e^ye^{x+y}}}\) 
Practice A19 
\(5x^22xy+7y^2=0\) 


\(\displaystyle{\frac{dy}{dx}=\frac{y5x}{7yx}}\) 
Practice A19 Final Answer 
\(\displaystyle{\frac{dy}{dx}=\frac{y5x}{7yx}}\) 
Practice A20 
\(\displaystyle{e^{x/y}=xy}\) 


\(\displaystyle{\frac{dy}{dx}=\frac{y(e^{x/y}y)}{xe^{x/y}y^2}}\) 
Practice A20 Final Answer 
\(\displaystyle{\frac{dy}{dx}=\frac{y(e^{x/y}y)}{xe^{x/y}y^2}}\) 
Practice A21 
\(x^27xy+3y^2=5\) 


\(\displaystyle{\frac{dy}{dx}=\frac{2x7y}{7x6y}}\) 
\(\displaystyle{
\begin{array}{rcl}
\frac{d}{dx}\left[ x^2  7xy + 3y^2 \right] & = & \frac{d}{dx}[ 5 ] \\
2x  7 \frac{d}{dx}[xy] + 3\frac{d}{dx}[y^2] & = & 0 \\
2x  7[ x \frac{dy}{dx} + y ] + 3 (2y) \frac{dy}{dx} & = & 0 \\
2x  7x \frac{dy}{dx}  7y + 6y \frac{dy}{dx} & = & 0 \\
7x \frac{dy}{dx} + 6y \frac{dy}{dx} & = & 7y  2x \\
(6y  7x) \frac{dy}{dx} & = & \\
\frac{dy}{dx} & = & \frac{7y2x}{6y7x}
\end{array}}\)
Practice A21 Final Answer 
\(\displaystyle{\frac{dy}{dx}=\frac{2x7y}{7x6y}}\) 
Practice A22 
Find all points on the curve \(x^{2/3}+y^{2/3}=8\) where the tangent line has slope \(m=1\). 


\((8,8), (8,8)\) 
Practice A22 Final Answer 
\((8,8), (8,8)\) 
Practice A23 
\(\displaystyle{\frac{x}{xy}=y^21}\) 


\(\displaystyle{\frac{dy}{dx}=\frac{y}{x2y(xy)^2}}\) 
Practice A23 Final Answer 
\(\displaystyle{\frac{dy}{dx}=\frac{y}{x2y(xy)^2}}\) 
Practice B01 
\(x^2+xy+\cos(y)=8y\) 


\(\displaystyle{\frac{dy}{dx}=\frac{(2x+y)}{x\sin(y)8}}\) 
Practice B01 Final Answer 
\(\displaystyle{\frac{dy}{dx}=\frac{(2x+y)}{x\sin(y)8}}\) 
Practice B02 
\(x^3xy+y^2=13e^y\) 


\(\displaystyle{\frac{dy}{dx}=\frac{y3x^2}{x+2y13e^y}}\) 
Find \(dy/dx\) for \(x^3xy+y^2=13e^y\).
Notice we can't solve for y here, so we need to use implicit differentiation.
This looks more difficult than it really is. We will break down each step and, when you are done going through it (which you may need to do several times to completely understand it), you will be able to work most any implicit differentiation problem. We will lay out all the steps and then explain each one in detail.
First, we take the derivative of both sides of the equation. 
\(\displaystyle{ \frac{d}{dx} \left[ x^3  xy + y^2 \right] }\) 
\(=\) 
\(\displaystyle{ \frac{d}{dx}[ 13e^y ] }\) 
On the left of the equal sign, we use the addition and subtraction rules to take the derivative of each term. 
\(\displaystyle{ \frac{d}{dx}[x^3]  \frac{d}{dx}[xy] + \frac{d}{dx}[y^2] }\) 
\(=\) 
\(\displaystyle{ \frac{d}{dx}[13e^y] }\) 
To explain what we do next, we will look at each term individually.
\(\displaystyle{ \frac{d}{dx}[x^3] ~~~~~ \to ~~~~~ 3x^2 }\)    for this term, we use the power rule.
\(\displaystyle{ \frac{d}{dx}[xy] ~~~~~ \to ~~~~~ \left[ x(dy/dx) + y(1) \right] }\)    here, we use the product rule on the entire term. When we get to \( d[y]/dx \), the chain rule tells us that since \(y\) does not match the \(x\) in the \(dx\) of the derivative, we cannot take the derivative of directly. So we are left with the term \( d[y]/dx \). This is written in the shorthand version \( dy/dx \).
The third term on the left of the equal sign is \(\displaystyle{ \frac{d}{dx}[y^2] ~~~~~ \to ~~~~~ 2y (dy/dx) }\). We use the chain rule again here.
Now, the right side of the equal sign is similar. We pull out the constant using the constant multiple rule and take the derivative as follows
\(\displaystyle{ \frac{d}{dx}[ 13e^y ] = 13 \frac{d}{dx}[ e^y ] = 13e^y (dy/dx) }\). Again, the chain rule is required giving us the term \(dy/dx\). 
\(\displaystyle{ 3x^2  \left[ x \frac{dy}{dx} + y(1) \right] + 2y \frac{dy}{dx} }\) 
\(=\) 
\(\displaystyle{ 13e^y \frac{dy}{dx} }\) 
In this step, we just clean up using algebra, being very careful to distribute the negative sign on the left. 
\(\displaystyle{ 3x^2  x \frac{dy}{dx}  y +2y \frac{dy}{dx} }\) 
\(=\) 
\(\displaystyle{ 13e^y \frac{dy}{dx} }\) 
Now our goal is to solve for \(dy/dx\). In this step, we move everything without \(dy/dx\) to the right and everything with \(dy/dx\) to the left. In this case, need to add \(y  3x^2\) to both sides and subtract \(13e^y(dy/dx)\) from both sides. 
\(\displaystyle{ x \frac{dy}{dx} + 2y \frac{dy}{dx}  13e^y \frac{dy}{dx} }\) 
\(=\) 
\(\displaystyle{ y  3x^2 }\) 
Okay, so in this line, we factor out the \(dy/dx\) from each term on the left. 
\(\displaystyle{ (x + 2y  13e^y)\frac{dy}{dx} }\) 
\(=\) 
\(\displaystyle{ y  3x^2 }\) 
Finally, we divide both sides by \((x + 2y  13e^y)\), to solve for \(dy/dx\). 
\(\displaystyle{ \frac{dy}{dx} }\) 
\(=\) 
\(\displaystyle{ \frac{y  3x^2}{x + 2y  13e^y} }\) 
Practice B02 Final Answer 
\(\displaystyle{\frac{dy}{dx}=\frac{y3x^2}{x+2y13e^y}}\) 
Practice B03 
\(\displaystyle{e^{(2x+3y)}=x^2\ln(xy^3)}\) 


\(\displaystyle{dy/dx=\frac{y\left[2x^212xe^{(2x+3y)}\right]}{3x\left[1+ye^{(2x+3y)}\right]}}\) 
First, we expand the right side of the original equation. This is not necessary but it will make the derivative of the natural log term easier to evaluate.
\( \ln(xy^3) = \ln(x) + \ln(y^3) = \ln(x) + 3\ln(y)\)
We will lay out the solution stepbystep and explain each line individually.
take the derivative of both sides of the equation 
\(\displaystyle{\frac{d}{dx} \left[ e^{(2x+3y)} \right]}\) 
\(=\) 
\(\displaystyle{\frac{d}{dx} \left[ x^2  \ln(x)  3\ln(y) \right]}\) 
On the left side of the equal sign and on the last term on the right, use the chain rule. 
\(\displaystyle{e^{(2x+3y)} \frac{d}{dx}[2x+3y]}\) 
\(=\) 
\(\displaystyle{2x  \frac{1}{x}  3 \frac{1}{y} \frac{dy}{dx}}\) 
\(\displaystyle{e^{(2x+3y)} \left[ 2 + 3 \frac{dy}{dx} \right]}\)  \(=\) 
\(\displaystyle{2x  \frac{1}{x}  \frac{3}{y} \frac{dy}{dx}}\) 
On the left, distribute the exponential term in anticipation of solving for \(dy/dx\). 
\(\displaystyle{2 e^{(2x+3y)} + 3e^{(2x+3y)} \frac{dy}{dx}}\)  \(=\)  
Move all terms with \(dy/dx\) to the left and other terms to the right of the equal sign. 
\(\displaystyle{3e^{(2x+3y)} \frac{dy}{dx} + \frac{3}{y} \frac{dy}{dx}}\)  \(=\) 
\(\displaystyle{2x  \frac{1}{x}  2 e^{(2x+3y)}}\) 
Factor out the \(dy/dx\) on the left and, on both sides, get a common denominator. 
\(\displaystyle{\left[ \frac{3ye^{(2x+3y)}}{y} + \frac{3}{y} \right] \frac{dy}{dx}}\)  \(=\) 
\(\displaystyle{\frac{2x^2  1  2x e^{(2x+3y)} }{x}}\) 
Solve for for \(dy/dx\). 
\(\displaystyle{\frac{dy}{dx}}\)  \(=\) 
\(\displaystyle{\frac{y \left[ 2x^2  1  2xe^{(2x+3y)} \right]}{3x \left[ 1 + y e^{(2x+3y)} \right]}}\) 
Practice B03 Final Answer 
\(\displaystyle{dy/dx=\frac{y\left[2x^212xe^{(2x+3y)}\right]}{3x\left[1+ye^{(2x+3y)}\right]}}\) 
Practice B04 
\(\displaystyle{\cos(x^2+2y)=xe^{y^2}}\) 


\(\displaystyle{\frac{dy}{dx}=\frac{e^{y^2}+2x\sin(x^2+2y)}{2\left[\sin(x^2+2y)+xye^{y^2}\right]}}\) 
Take the derivative of both sides. 
\(\displaystyle{ \frac{d}{dx} \left[ \cos(x^2+2y) \right] }\) 
\(=\) 
\(\displaystyle{ \frac{d}{dx} \left[ xe^{y^2} \right] }\) 
Use the chain rule on the left and the product rule on the right. 
\(\displaystyle{ \sin(x^2+2y) \frac{d}{dx}[x^2+2y] }\) 
\(=\) 
\(\displaystyle{ x \frac{d}{dx}[e^{y^2}] + e^{y^2} (1) }\) 
Use the chain rule on the \(2y\) term on the left and again on the exponential term on the right. 
\(\displaystyle{ \sin(x^2+2y) \left[ 2x + 2 dy/dx \right] }\) 
\(=\) 
\(\displaystyle{ x e^{y^2} (2y)dy/dx + e^{y^2} }\) 
On the left, distribute the sine term. On the right, simplify the first term. 
\(\displaystyle{ 2x \sin(x^2+2y)  2\sin(x^2+2y)dy/dx }\) 
\(=\) 
\(\displaystyle{ 2xy e^{y^2} dy/dx + e^{y^2} }\) 
Move all terms with \(dy/dx\) to the left of the equal sign and all terms without the derivative term to the right. 
\(\displaystyle{  2\sin(x^2+2y)dy/dx  2xy e^{y^2} dy/dx }\) 
\(=\) 
\(\displaystyle{ e^{y^2} + 2x \sin(x^2+2y) }\) 
On the left, factor out \(dy/dx\). 
\(\displaystyle{ \left[  2\sin(x^2+2y)  2xy e^{y^2} \right] dy/dx }\) 
\(=\) 
\(\displaystyle{ }\) 
Solve for \(dy/dx\). 
\(\displaystyle{ \frac{dy}{dx} }\) 
\(=\) 
\(\displaystyle{ \frac{e^{y^2} + 2x \sin(x^2+2y)}{ 2\sin(x^2+2y)  2xy e^{y^2} } }\) 
Finally, factor out the \(2\) from each term in the denominator and write the negative sign in front of the fraction. 
\(\displaystyle{ \frac{dy}{dx} }\) 
\(=\) 
\(\displaystyle{  \frac{e^{y^2} + 2x \sin(x^2+2y)}{2\left[ \sin(x^2+2y) + xy e^{y^2} \right]} }\) 
Practice B04 Final Answer 
\(\displaystyle{\frac{dy}{dx}=\frac{e^{y^2}+2x\sin(x^2+2y)}{2\left[\sin(x^2+2y)+xye^{y^2}\right]}}\) 
Practice B05 
\(\displaystyle{\tan(x^2y^4)=3x+y^2}\) 


\(\displaystyle{\frac{dy}{dx}=\frac{32xy^4\sec^2(x^2 y^4)}{2y(2x^2y^2\sec^2(x^2y^4)1)}}\) 
\(\displaystyle{
\begin{array}{rcl}
\frac{d}{dx}[\tan(x^2 y^4)] & = & \frac{d}{dx} [3x+y^2] \\
\frac{d}{dx}[x^2 y^4] \sec^2 (x^2 y^4) & = & 3 + 2y \frac{dy}{dx} \\
\left[ x^2 (4y^3)\frac{dy}{dx} + y^4 (2x) \right] \sec^2 (x^2 y^4) & = & \\
4x^2y^3 \sec^2 (x^2 y^4) \frac{dy}{dx} + 2xy^4 \sec^2 (x^2 y^4) & = & \\
4x^2y^3 \sec^2 (x^2 y^4) \frac{dy}{dx}  2y \frac{dy}{dx} & = & 3  2xy^4 \sec^2 (x^2 y^4) \\
\left[ 4x^2y^3 \sec^2 (x^2 y^4)  2y \right] \frac{dy}{dx} & = & \\
\frac{dy}{dx} & = & \frac{3  2xy^4 \sec^2 (x^2 y^4)}{2y(2x^2y^2 \sec^2 (x^2 y^4)  1)}
\end{array}}\)
Practice B05 Final Answer 
\(\displaystyle{\frac{dy}{dx}=\frac{32xy^4\sec^2(x^2 y^4)}{2y(2x^2y^2\sec^2(x^2y^4)1)}}\) 
Practice B06 
\(\sec(xy)+e^y=3x4\) 


\(\displaystyle{\frac{dy}{dx}=\frac{3y\sec(xy)\tan(xy)}{x\sec(xy)\tan(xy)+e^y}}\) 
Practice B06 Final Answer 
\(\displaystyle{\frac{dy}{dx}=\frac{3y\sec(xy)\tan(xy)}{x\sec(xy)\tan(xy)+e^y}}\) 
Practice B07 
\(\displaystyle{6xy^{1/2}+6y=72}\); find \(dy/dx\) and slope at \((4,4)\) 


\(\displaystyle{\frac{dy}{dx}=\frac{2y}{x+2y^{1/2}}}\) the slope at the point \((4,4)\) is \(m=1\) 
First, we will find the equation for \(dy/dx\).
\(\displaystyle{
\begin{array}{rcl}
\frac{d}{dx}\left[ 6xy^{1/2} + 6y \right] & = & \frac{d}{dx}[72] \\
6x \frac{d}{dx}[y^{1/2}] + 6y^{1/2} (1) + 6 \frac{dy}{dx} & = & 0 \\
6x (1/2)y^{1/2} \frac{dy}{dx} + 6y^{1/2} + 6 \frac{dy}{dx} & = & 0 \\
\frac{x}{2y^{1/2}} \frac{dy}{dx} + \frac{dy}{dx} & = & y^{1/2} \\
\left[ \frac{x}{2y^{1/2}} + 1 \right] \frac{dy}{dx} & = & \\
\left[ \frac{x+2y^{1/2}}{2y^{1/2}} \right] \frac{dy}{dx} & = & \\
\frac{dy}{dx} & = & \frac{(y^{1/2}) (2 y^{1/2})}{x+2y^{1/2}} \\
& = & \frac{2y}{x+2y^{1/2}}
\end{array}}\)
So, now we have an expression for the derivative everywhere on the graph. We need to find the slope at the point \((4,4)\). But, first, let's doublecheck that the point is on the graph. We do that by plugging the point into the original equation.
\(\displaystyle{
\begin{array}{rcl}
6(4)(4)^{1/2} + 6(4) & = & 72 \\
24(2) + 24 & = & 72 \\
72 & = & 72 √
\end{array}}\)
Check.
So, to calculate the slope at the point, we plug the point into the expression for the slope.
\(\displaystyle{\frac{dy}{dx} = \frac{2y}{x+2y^{1/2}} ~~~ \to ~~~ \frac{dy}{dx} = \frac{2(4)}{4+2{4}^{1/2}} = 1}\)
Practice B07 Final Answer 
\(\displaystyle{\frac{dy}{dx}=\frac{2y}{x+2y^{1/2}}}\) the slope at the point \((4,4)\) is \(m=1\) 
Practice B08 
Find \( y'' \) for \(9x^2+y^2=9\) 


\(\displaystyle{\frac{d^2y}{dx^2}=\frac{81}{y^3}}\) 
Practice B08 Final Answer 
\(\displaystyle{\frac{d^2y}{dx^2}=\frac{81}{y^3}}\) 
Practice B09 
\(x^2+y^2=(2x^2+2y^2x)^2\); find \(dy/dx\) and the tangent line at \((0,1/2)\) 


\(\displaystyle{dy/dx=\frac{8x^36x^2+8xy^22y^2}{y8x^2y8y^3+4xy}}\) the equation of the tangent line at the point \((0,1/2)\) is \(y=x+1/2\) 
Practice B09 Final Answer 
\(\displaystyle{dy/dx=\frac{8x^36x^2+8xy^22y^2}{y8x^2y8y^3+4xy}}\) the equation of the tangent line at the point \((0,1/2)\) is \(y=x+1/2\) 
Practice B10 
\(\displaystyle{xy+e^y=e}\); find the second derivative at \(x=0\) 


\(\displaystyle{e^{2}}\) 
Practice B10 Final Answer 
\(\displaystyle{e^{2}}\) 
Practice B11 
find \( y''\) of \(x^4+y^4=16\) 


\(\displaystyle{\frac{d^2y}{dx^2}=\frac{3x^2(y^4+x^4)}{y^7}=\frac{48x^2}{y^7}}\) 
Practice B11 Final Answer 
\(\displaystyle{\frac{d^2y}{dx^2}=\frac{3x^2(y^4+x^4)}{y^7}=\frac{48x^2}{y^7}}\) 
Practice B12 
\(\cos x^2+\sin y^3=0.6\) 


\(\displaystyle{\frac{dy}{dx}=\frac{2x\sin(x^2)}{3y^2\cos(y^3)}}\) 
Practice B12 Final Answer 
\(\displaystyle{\frac{dy}{dx}=\frac{2x\sin(x^2)}{3y^2\cos(y^3)}}\) 
Practice B13 
\(1/x+\ln y+(xy)^3=4x\) 


\(\displaystyle{\frac{dy}{dx}=\frac{4+1/x^23x^2y^3}{1/y+3x^3y^2}}\) 
Practice B13 Final Answer 
\(\displaystyle{\frac{dy}{dx}=\frac{4+1/x^23x^2y^3}{1/y+3x^3y^2}}\) 