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You CAN Ace Calculus

17calculus > derivatives > implicit differentiation

 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

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 [6min-47secs]

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 [2min-52secs]

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 $$y=x^2+5$$

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 y-term.

 Take the derivative of $$y^2$$ with respect to $$x$$.
 Use the result from the previous example, to find the derivative of $$e^{y^2}$$ with respect to $$x$$.

Okay, so now you are ready to see how we use this technique to determine $$dy/dx$$.

 Find $$dy/dx$$ of $$y^2 - x = 7$$.

Another term you will run across when doing implicit differentiation is $$xy$$.

 Take the derivative of $$xy$$ with respect to $$x$$.
 Use the result from the previous example to find the derivative of $$\sin(xy)$$ with respect to $$x$$.

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.

### Search 17Calculus

Practice Problems

Instructions - - Unless otherwise instructed, calculate dy/dx of the following equations using implicit differentiation. Assume y is a function of x.
- When you are asked to find the equation of tangent lines, give your answers in slope-intercept 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.

 Level A - Basic

Practice A01

$$y=e^{xy}$$

solution

Practice A02

$$x^2+y^2=25$$; find $$dy/dx$$ and the slope at $$x=-3$$

solution

Practice A03

$$2x^3+y^3-7=0$$

solution

Practice A04

For $$xy^2=100$$, calculate $$dy/dx$$, the tangent line at $$(1,10)$$ and graph.

solution

Practice A05

$$\sin(x/y)=1/2$$

solution

Practice A06

$$(x+y)^2=x-y-3$$

solution

Practice A07

For $$(x-y-1)^3=x$$, find $$dy/dx$$, the tangent line at $$(1,-1)$$ and graph.

solution

Practice A08

For $$e^{xy}+x^2=y^2$$, calculate $$dy/dx$$, the tangent line at $$(0,1)$$ and graph.

solution

Practice A09

find tangent line of $$y\sin(2x)=x\cos(2y)$$ at $$(\pi/2,\pi/4)$$.

solution

Practice A10

$$1+x=\sin(xy^2)$$

solution

Practice A11

$$\displaystyle{x^2+xy+y^2=3}$$; find $$dy/dx$$ and the tangent line at $$(1,1)$$

solution

Practice A12

$$x^3+y^3=xy$$

solution

Practice A13

$$x^2y+xy^2=3x$$

solution

Practice A14

$$\displaystyle{x^{-2}-\ln(y^3)=7}$$

solution

Practice A15

find $$ds/dt$$ for $$s^3-t^3=8$$

solution

Practice A16

find $$dy/dt$$ for $$y^3+\sin(yt)=t^5$$

solution

Practice A17

$$x\sin(y)+y\sin(x)=1$$

solution

Practice A18

$$\displaystyle{2x+3e^y=e^{x+y}}$$

solution

Practice A19

$$5x^2-2xy+7y^2=0$$

solution

Practice A20

$$\displaystyle{e^{x/y}=x-y}$$

solution

Practice A21

$$x^2-7xy+3y^2=5$$

solution

Practice A22

Find all points on the curve $$x^{2/3}+y^{2/3}=8$$ where the tangent line has slope $$m=-1$$.

solution

Practice A23

$$\displaystyle{\frac{x}{x-y}=y^2-1}$$

solution

 Level B - Intermediate

Practice B01

$$x^2+xy+\cos(y)=8y$$

solution

Practice B02

$$x^3-xy+y^2=13e^y$$

solution

Practice B03

$$\displaystyle{e^{(2x+3y)}=x^2-\ln(xy^3)}$$

solution

Practice B04

$$\displaystyle{\cos(x^2+2y)=xe^{y^2}}$$

solution

Practice B05

$$\displaystyle{\tan(x^2y^4)=3x+y^2}$$

solution

Practice B06

$$\sec(xy)+e^y=3x-4$$

solution

Practice B07

$$\displaystyle{6xy^{1/2}+6y=72}$$; find $$dy/dx$$ and slope at $$(4,4)$$

solution

Practice B08

Find $$y''$$ for $$9x^2+y^2=9$$

solution

Practice B09

$$x^2+y^2=(2x^2+2y^2-x)^2$$; find $$dy/dx$$ and the tangent line at $$(0,1/2)$$

solution

Practice B10

$$\displaystyle{xy+e^y=e}$$; find the second derivative at $$x=0$$

solution

Practice B11

find $$y''$$ of $$x^4+y^4=16$$

solution

Practice B12

$$\cos x^2+\sin y^3=0.6$$

$$1/x+\ln y+(xy)^3=4x$$