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Basic Trig Derivatives (no chain rule required) |
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This first section discuss the basics of trig derivatives and does not require you to know the chain rule. Here are the basic rules. If you work enough practice problems, you won't need to memorize them since you will just know them.
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
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\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
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\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
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\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
The first two, sine and cosine, are pretty straightforward since they are ALMOST inverses of each other. You just need to remember the negative sign on the second one.The rest of them can be derived from the sine and cosine rules using the product rule, quotient rule and basic trigonometric identities. Here is a good video showing this derivation.
| PatrickJMT: Deriving the Derivative Formulas for Tangent, Cotangent, Secant, Cosecant |
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Notice how the derivatives seem to be in similar pairs. Take a few minutes and look for patterns and similarities as well as differences in all three sets. Take special notice of when a minus sign appears.
Okay, let's take a minute and watch a quick video clip for another perspective on how to remember these derivatives.
| Krista King Math: Derivatives of TRIG FUNCTIONS |
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Here are some practice problems. Calculate the derivative of these functions.
Practice 1 |
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\(\displaystyle{x^4\tan(x)}\) |
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\(\displaystyle{\frac{d}{dx}[ x^4 \tan(x) ] = x^3[x \sec^2(x)+4 \tan(x)]}\) |
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\(\displaystyle{\frac{d}{dx}[ x^4 \tan(x) ] = x^4 \sec^2(x) + 4x^3 \tan(x) = }\) \(\displaystyle{x^3[x \sec^2(x)+4 \tan(x)]}\)
Note: Not factoring out the \(x^3\) term, may cost you points.
Practice 1 Final Answer |
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\(\displaystyle{\frac{d}{dx}[ x^4 \tan(x) ] = x^3[x \sec^2(x)+4 \tan(x)]}\) |
Practice 2 |
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\(\displaystyle{f(x)=\cos(x)\sin(x)}\) |
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\(\displaystyle{ \frac{d}{dx}\left[ \cos(x) \sin(x) \right] = \cos^2(x) - \sin^2(x) }\) |
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\(\displaystyle{
\begin{array}{rcl}
\frac{d}{dx}\left[ \cos(x) \sin(x) \right] & = & \cos(x) \frac{d}{dx}[ \sin(x)] + \sin(x) \frac{d}{dx}[ \cos(x)] \\
& = & \cos(x)[ \cos(x)] + \sin(x)[-\sin(x)] \\
& = & \cos^2(x) - \sin^2(x) \end{array}}\)
Practice 2 Final Answer |
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\(\displaystyle{ \frac{d}{dx}\left[ \cos(x) \sin(x) \right] = \cos^2(x) - \sin^2(x) }\) |
Practice 3 |
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\(\displaystyle{\frac{\sin(x)}{1+\cos(x)}}\) |
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\(\displaystyle{\frac{d}{dx}\left[\frac{\sin(x)}{1+\cos(x)}\right]=\frac{1}{1+\cos(x)}}\) |
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\(
\begin{array}{rcl}
& & \displaystyle{\frac{d}{dx}\left[ \frac{\sin(x)}{1+\cos(x)} \right]} \\
& = & \displaystyle{\frac{[1+\cos(x)]d[\sin(x)]/dx - [\sin(x)]d[1+\cos(x)]/dx}{[1+\cos(x)]^2}} \\
& = & \displaystyle{\frac{[1+\cos(x)][\cos(x)] - [\sin(x)][-\sin(x)]}{[1+\cos(x)]^2} } \\
& = & \displaystyle{\frac{\cos(x)+\cos^2(x)+\sin^2(x)}{[1+\cos(x)]^2} } \\
& = & \displaystyle{\frac{1+\cos(x)}{[1+\cos(x)]^2} } \\
& = & \displaystyle{ \frac{1}{1+\cos(x)} }
\end{array}\)
Notice that we used the identity \( \sin^2(x) + \cos^2(x) = 1 \). Whenever you have sines and cosines, watch for this to show up to help simplify.
Practice 3 Final Answer |
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\(\displaystyle{\frac{d}{dx}\left[\frac{\sin(x)}{1+\cos(x)}\right]=\frac{1}{1+\cos(x)}}\) |
Practice 4 |
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\(\displaystyle{y=\frac{\tan(x)}{x^{3/2}+5x}}\) |
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Practice 5 |
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\(\displaystyle{\frac{3}{x}\cot(x)}\) |
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\(\displaystyle{ \frac{d}{dx}\left[ \frac{3}{x} \cot(x) \right] = \frac{-3}{x^2}\left[ x \csc^2(x) + \cot(x) \right] }\) |
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\(
\begin{array}{rcl}
\displaystyle{\frac{d}{dx}\left[ \frac{3}{x} \cot(x) \right]}
& = & \displaystyle{ \frac{3}{x}\frac{d}{dx}[\cot(x)] + \cot(x) \frac{d}{dx}[3/x] } \\
& = & \displaystyle{ \frac{3}{x} (-\csc^2(x)) + \cot(x) \frac{d}{dx}[3x^{-1}] } \\
& = & \displaystyle{ \frac{-3}{x} \csc^2(x) + \cot(x) [3(-1)x^{-2}] } \\
& = & \displaystyle{ \frac{-3}{x} \csc^2(x) - \frac{3}{x^2} \cot(x) } \\
& = & \displaystyle{ \frac{-3}{x^2}[ x \csc^2(x) + \cot(x) ] }
\end{array}\)
Practice 5 Final Answer |
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\(\displaystyle{ \frac{d}{dx}\left[ \frac{3}{x} \cot(x) \right] = \frac{-3}{x^2}\left[ x \csc^2(x) + \cot(x) \right] }\) |
Basic Inverse Trig Derivatives (no chain rule required) |
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You might expect the derivatives for inverse trig functions to be similar to derivatives for trig functions but they are not. They are very different. But, again, they appear in similar pairs.
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
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\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
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\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
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\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Important Things To Notice
1. Strangely enough, the derivative of inverse trig functions do not contain any trig or inverse trig terms. We include some videos below showing the derivation of these equations.
2. Each pair of inverse trig derivatives are very closely related, even closer than with trig derivatives. Each pair is the same EXCEPT for a negative sign. So, for example, \( [\arccos(t)]' = -[\arcsin(t)]' \).
3. In the case of the third pair, \( [\arcsec(t)]' \) and \( [\arccsc(t)]' \), the denominators contain an absolute value term, \( \abs{t} \), which is important. Do not leave off the absolute value signs unless you explicitly state that \( t \) is always positive. Keeping them is always the safe way to make sure you are correct. ( Of course, you need to check with your instructor to see what they require. )
4. Remember that the notation \(\sin^{-1}(t)\) actually means \(\arcsin(t)\), not \(1/\sin(t)\).
Before we go on, let's watch some videos showing how the derivatives above come about. It seems strange that when we take the derivative of a function involving inverse trig functions, there are no longer any trig or inverse trig terms. These next videos show how to get the derivatives. Watching these will help you understand trig and inverse trig in more depth.
| PatrickJMT: The Derivative of Inverse Sine or y = arcsin(x) |
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| MIP4U: Proof - The Derivative of f(x)=arccos(x): d/dx[arccos(x)] |
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| PatrickJMT: Deriving the Derivative of Inverse Tangent or y = arctan (x) |
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| MIP4U: Proof - The Derivative of f(x)=arccot(x): d/dx[arccot(x)] |
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| MIP4U: Proof - The Derivative of f(x)=arcsec(x): d/dx[arcsec(x)] |
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| MIP4U: Proof - The Derivative of f(x)=arccsc(x): d/dx[arccsc(x)] |
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Trig Derivatives Using The Chain Rule |
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This section requires understanding of the chain rule and the previous sections on the basics of trig derivatives. Except for the most basic problems, you will find most problems require the use of the chain rule.
As we discuss on the chain rule page, we think the easiest way to work problems that require the chain rule is to start on the outside and work your way in. Doing it the other direction, requires you to know where to start and what is actually on the inside. Especially when starting out, this is more complicated in our opinion. ( Of course, you need to consult your instructor to find out what they require. )
The best way to work these derivatives is to use substitution, especially while you are still learning. Later, you can do it in your head. In the examples on this page, we will write out the substitution to help you see it.
Let's start out by watching a video. This is a great video explaining the idea of the chain rule when a trig function is involved.
| Krista King Math: Chain rule for derivatives, with trig functions |
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Okay, let's work some examples. If you feel up to it, try to work these before opening the solutions.
| Example 1 - - Evaluate \( [\sin(x^2)]' \) |
We can tell that the outside function is sine. So, we use the substitution \( u = x^2 \). The reason we chose \( x^2 \) is because we want \( u \) to be equal to everything inside the sine function.
\(\displaystyle{
\begin{array}{rcl}
\frac{d}{dx}[\sin(x^2)] & = & \frac{d}{dx}[\sin(u)] \\
& = & \frac{d}{du}[\sin(u)] \frac{d}{dx}[u] \\
& = & \frac{d}{du}[\sin(u)] \frac{d}{dx}[x^2] \\
& = & \cos(u) (2x) \\
& = & 2x \cos(x^2)
\end{array}
}\)
Notes:
1. The step between lines 1 and 2 above is where we applied the chain rule.
2. Once we have taken all derivatives, the second to the last line is \( \cos(u)(2x) \). We can't leave the answer in this form. We need to reverse the substitution of \( u \). Remember, the substitution technique is used only temporarily during computations and none of the temporary variables should appear in the final answer.
So our final answer is \( [\sin(x^2)]' = 2x \cos(x^2) \).
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| Example 2 - - Evaluate \( [\sec(x^2)]' \) |
The outside function is secant. So, we will again use the substitution \( u = x^2 \).
\(\displaystyle{
\begin{array}{rcl}
\frac{d}{dx}[\sec(x^2)] & = & \frac{d}{dx}[\sec(u)] \\
& = & \frac{d}{du}[\sec(u)] \frac{d}{dx}[u] \\
& = & \frac{d}{du}[\sec(u)] \frac{d}{dx}[x^2] \\
& = & \sec(u) \tan(u) (2x) \\
& = & 2x \sec(x^2) \tan(x^2)
\end{array}
}\)
Note:
There is really nothing new here that wasn't in the first example, except that, at the end, there is more than one place where we need to substitute \( u \) back in, i.e. \( \sec(u) ~~ \to ~~ \sec(x^2) \) and \( \tan(u) ~~ \to ~~ \tan(x^2) \).
So our final answer is \( [\sec(x^2)]' = 2x \sec(x^2) \tan(x^2) \).
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This next example uses the product rule and derivative of exponentials. If you have not studied both of these topics, you can skip this example and come back to it once you have. We will also show you how to work a nested chain rule problem.
| Example 3 - - Evaluate \( [\csc(2xe^{15x})]' \) |
Okay, as before, we use substitution, letting \(u\) be equal to everything inside the cosecant term, i.e. \( u=2xe^{15x} \).
\(\displaystyle{
\begin{array}{rcl}
\frac{d}{dx}[\csc(2xe^{15x})] & = & \frac{d}{dx}[\csc(u)] \\
& = & \frac{d}{du}[\csc(u)] \frac{d}{dx}[u] \\
& = & -\csc(u) \cot(u) \frac{d}{dx}[2xe^{15x}] \\
& = & -\csc(u) \cot(u) \left[(2x) \frac{d}{dx}[e^{15x}] + e^{15x} \frac{d}{dx}[2x] \right] \\
& = & -\csc(u) \cot(u) \left[(2x) (e^{15x}) \frac{d}{dx}[15x] + e^{15x} (2) \right] \\
& = & -\csc(u) \cot(u) \left[(2x) (e^{15x}) (15) + e^{15x} (2) \right] \\
& = & -2 (15x+1) e^{15x} \csc(2xe^{15x}) \cot(2xe^{15x})
\end{array}
}\)
This looks somewhat complicated, so we will break it down for you.
The first two lines are as before, using substitution, we apply the chain rule. Looking at only the second term in line three, we have
\( \displaystyle{ \frac{d}{dx}[2xe^{15x}] }\) to which we apply the product rule, where the first term is \(2x\) and the second term is \(e^{15x}\). When we do, we get the fourth row above, the second part of which is
\( \displaystyle{ \frac{d}{dx}[2xe^{15x}] = 2x \frac{d}{dx}[e^{15x}] + e^{15x} \frac{d}{dx}[2x] }\)
Now we need to apply the chain rule again to the term \(\displaystyle{ \frac{d}{dx}[e^{15x}] }\) giving us \( 15e^{15x}\).
After that the rest is factoring and simplifying. So our final answer is
\(\displaystyle{ [\csc(2xe^{15x})]' = -2 (15x+1) e^{15x} \csc(2xe^{15x}) \cot(2xe^{15x}) }\)
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Here are some practice problems. Unless otherwise instructed, calculate the derivative of these functions.
Practice 6 |
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\(\displaystyle{ 7 \sin(x^2+1) }\) |
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\(\displaystyle{ \frac{d}{dx}[ 7 \sin(x^2+1) ] = (14x) \cos(x^2+1) }\) |
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\(\displaystyle{
\begin{array}{rcl}
\frac{d}{dx}[ 7\sin(x^2+1) ] & = & 7\frac{d}{dx}[\sin(x^2+1)] \\\\
& = & 7\cos(x^2+1)\frac{d}{dx}[x^2+1] \\\\
& = & 7\cos(x^2+1)(2x) \\\\
& = & (14x)\cos(x^2+1)
\end{array}
}\)
In the first line above, we used the Constant Multiple Rule to move the 7 outside the derivative. Then we used the Chain Rule. A more explicit way to work this is to use substitution. Let \( u = x^2+1 \) which is the inside term. Then we would write
\(\displaystyle{
\begin{array}{rcl}
\frac{d}{dx}[ 7\sin(x^2+1) ] & = & 7\frac{d}{du}[\sin(u)] \cdot \frac{d}{dx}[x^2+1] \\\\
& = & 7[\cos(u)](2x) \\\\
& = & (14x)\cos(x^2+1)
\end{array}
}\)
If we call the original function, say \(f(x)=7\sin(x^2+1)\), the first line is \(\displaystyle{ \frac{df}{dx} = \frac{df}{du}\cdot\frac{du}{dx} }\) using the Chain Rule. This is probably a better way to work the problem than the first solution, especially while you are learning. Once you have the chain rule down, you can easily work it as shown in the first solution.
Practice 6 Final Answer |
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\(\displaystyle{ \frac{d}{dx}[ 7 \sin(x^2+1) ] = (14x) \cos(x^2+1) }\) |
Practice 7 |
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\(\displaystyle{ y=\sin^3(x) \tan(4x) }\) |
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Practice 8 |
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\(\displaystyle{ y = (2x^5-3) \sin(7x) }\) |
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\(\displaystyle{ \frac{dy}{dx} = 7(2x^5-3)\cos(7x) + (10x^4)\sin(7x) }\) |
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\(
\begin{array}{rcl}
\frac{dy}{dx} & = & (2x^5-3) \frac{d}{dx}[\sin(7x)] + \sin(7x)\frac{d}{dx} [2x^5-3] \\\\
& = & (2x^5-3)\cos(7x) (7) + \sin(7x) (10x^4) \\\\
& = & 7(2x^5-3)\cos(7x) + (10x^4)\sin(7x)
\end{array}
\)
Practice 8 Final Answer |
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\(\displaystyle{ \frac{dy}{dx} = 7(2x^5-3)\cos(7x) + (10x^4)\sin(7x) }\) |
Practice 9 |
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\( \tan(x^5 + 2x^3 - 12x) \) |
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\( (5x^4 + 6x^2 - 12) \sec^2(x^5 + 2x^3 - 12x) \) |
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Using substitution, the outside function is tangent and the inside function is everything inside the tangent parentheses. So, we let \( u = x^5 + 2x^3 - 12x \).
\(\displaystyle{
\begin{array}{rcl}
\displaystyle{\frac{d}{dx}[\tan(x^5 + 2x^3 - 12x)]} & = & \displaystyle{\frac{d}{dx}[\tan(u)]} \\
& = & \displaystyle{\frac{d}{du}[\tan(u)] \frac{d}{dx}[u] } \\
& = & \displaystyle{\frac{d}{du}[\tan(u)] \frac{d}{dx}[x^5 + 2x^3 - 12x] } \\
& = & \sec^2(u) (5x^4 + 6x^2 - 12) \\
& = & (5x^4 + 6x^2 - 12) \sec^2(x^5 + 2x^3 - 12x)
\end{array}
}\)
Practice 9 Final Answer |
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\( (5x^4 + 6x^2 - 12) \sec^2(x^5 + 2x^3 - 12x) \) |
Practice 10 |
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\(\displaystyle{ y = (1+\cos^2(7x))^3 }\) |
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Practice 11 |
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\(y=\cot^4(x/2)\) |
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\(y'=-2\cot^3(x/2)\csc^2(x/2)\) |
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Practice 11 Final Answer |
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\(y'=-2\cot^3(x/2)\csc^2(x/2)\) |
Practice 12 |
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Calculate the third derivative of \(f(x)=\tan(3x)\). |
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\(54\sec^2(3x)\left[ 2\tan^2(3x)+\sec^2(3x)\right]\) |
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Intermediate answers:
\(f'(x)=3\sec^2(3x)\)
\(f''(x)=18\sec^2(3x)\tan(3x)\)
Practice 12 Final Answer |
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\(54\sec^2(3x)\left[ 2\tan^2(3x)+\sec^2(3x)\right]\) |
Practice 13 |
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\(\displaystyle{ y=\left[ \tan \left( \sin \left( \sqrt{x^2+8x}\right) \right) \right]^5 }\) |
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Practice 14 |
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\(\displaystyle{ y=x \sin(1/x) + \sqrt[4]{(1-3x)^2 + x^5} }\) |
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In this video, his second term contains t of x.
Inverse Trig Derivatives Using The Chain Rule |
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As mentioned in the previous section, except for the simplest problems, you will need to use the chain rule for most inverse trig derivatives. Let's look at an example that parallels some of the examples in the previous section.
| Example 4 - - Evaluate \( [\arcsin(x^2)]' \) |
For this one, we let \( u = x^2 \). We chose \(x^2\) since we want \(u\) to be equal to everything inside the arcsine function.
\(\displaystyle{
\begin{array}{rcl}
\frac{d}{dx}[\arcsin(x^2)] & = & \frac{d}{dx}[\arcsin(u)] \\
& = & \frac{d}{du}[\arcsin(u)] \frac{d}{dx}[u] \\
& = & \frac{d}{du}[\arcsin(u)] \frac{d}{dx}[x^2] \\
& = & \frac{1}{\sqrt{1-u^2}} (2x) \\
& = & \frac{1}{\sqrt{1-(x^2)^2}} (2x) \\
& = & \frac{2x}{\sqrt{1-x^4}}
\end{array}
}\)
Notes:
1. The step between lines 1 and 2 is where we applied the chain rule.
2. Notice how, when we reverse substituted \( u = x^2 \), we had \( (x^2)^2 = x^4 \). This may be easily missed if you tried to do it in your head.
Our final answer is \(\displaystyle{ [\arcsin(x^2)]' = \frac{2x}{\sqrt{1-x^4}} }\)
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