Derivatives of Inverse Trig Functions

Suppose we want to find the derivative of $f^{-1}(x)$ for some function $f(x)$. Well, from Derivative of Inverse, we know $\frac{\text d}{\text dx}f^{-1}(x)=\frac{1}{f'(f^{-1}(x))}$.

Some Special Inverses

1. $\sin^{-1}(x)$, $\cos^{-1}(x)$, $\tan^{-1}(x)$

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$$\frac{\text d\sin^{-1}(x)}{\text dx}=\frac{1}{\cos(\sin^{-1}(x))}=\frac{1}{\sqrt{1-\sin^2(\sin^{-1}(x))}}=\frac{1}{\sqrt{1-x^2}}$$ $$\frac{\text d\cos^{-1}(x)}{\text dx}=\frac1{-\sin(\cos^{-1}(x))}=-\frac1{\sqrt{1-\cos^2(\cos^{-1}(x))}}=-\frac{1}{\sqrt{1-x^2}}$$ $$\frac{\text d\tan^{-1}(x)}{\text dx}=\frac1{\sec^2(\tan^{-1}(x))}=\frac1{1+\tan^2(\tan^{-1}(x))}=\frac{1}{1+x^2}$$

1. $\ln (x)$

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$$\frac{\text d\ln(x)}{\text dx}=\frac{1}{e^{\ln x}}=\frac1x$$

1. $\cosh^{-1}(x)$, $\sinh^{-1}(x)$

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$$\frac{\text d\cosh^{-1}(x)}{\text dx}=\frac{1}{\sinh(\cosh^{-1}(x))}=\frac1{\sqrt{\cosh^2(\cosh^{-1}(x))-1}}=\frac{1}{\sqrt{x^2-1}}$$ $$\frac{\text d\sinh^{-1}(x)}{\text dx}=\frac{1}{\cosh(\sinh^{-1}(x))}=\frac1{\sqrt{\sinh^2(\sinh^{-1}(x))+1}}=\frac1{\sqrt{x^2+1}}$$

Alternate proof: Proof by implicit differentiation

$\frac{\text{d}}{\text{d}x}(\sin^{-1}{x})$: We let $y=\sin^{-1}{x}$, so then $x=\sin{y}$. Thus $\frac{\text{d}}{\text{d}x}x=\frac{\text{d}}{\text{d}x}\sin(y)\rightarrow 1=\cos{y}\frac{\text{d}y}{\text{d}x} \rightarrow \frac{\text{d}y}{\text{d}x}=\frac{1}{\cos{y}}=\frac{1}{\sqrt{1-\sin^2{y}}}=\frac{1}{\sqrt{1-x^2}}$.

For $\cos^{-1}{x}$, we use the fact that $\cos^{-1}{x}+\sin^{-1}{x}=\frac{\pi}{2}$, which is a constant. Thus $\frac{\text{d}}{\text{d}x}\cos^{-1}{x} + \frac{\text{d}}{\text{d}x}\sin^{-1}{x}=0$, so $\frac{\text{d}}{\text{d}x}\cos^{-1}{x}=-\frac{1}{\sqrt{1-x^2}}$.

For $\tan^{-1}{x}$, we let $x=\tan{y}$ so $1=\sec^{2}{y}\frac{\text{d}y}{\text{d}x}$, so $\frac{\text{d}y}{\text{d}x}=\frac{1}{\sec^{2}{y}}=\frac{1}{1+\tan^{2}{y}}=\frac{1}{1+x^2}$.

The derivative of $\cot^{-1}{x}$ is $-\frac{1}{1+x^2}$ because of the inverse cofunction identities.

With $\sec^{-1}{x}$, we again let $x=\sec{y}$. Now we have $1=\sec{y}\tan{y}\frac{\text{d}y}{\text{d}x}\rightarrow \frac{\text{d}y}{\text{d}x}=\frac{1}{\sec{y}\tan{y}}=\pm \frac{1}{\sec{y}\sqrt{\sec^{2}{y}-1}} = \pm \frac{1}{x\sqrt{x^2-1}}$. Now we notice that the sign is positive if $x \ge 1$ and negative if $x \le -1$, so we simply have that $\frac{\text{d}y}{\text{d}x} = \frac{1}{|x|\sqrt{x^2-1}}$. The derivative for $\csc^{-1}{x}$ is again the negative of that.

For $\ln{x}$: $x = e^{y} \rightarrow \frac{\text{d}y}{\text{d}x} \cdot e^y = 1 \rightarrow \frac{\text{d}y}{\text{d}x} = \frac{1}{e^{y}}=\frac{1}{x}$.

I might as well note that since $\log_a{x}=\frac{\ln{x}}{\ln{a}}$, its derivative is simply $\frac{1}{x\ln{a}}$.

Here is some practice with taking derivatives of inverse functions, good luck!

(Videos) Derivatives of Inverse Trig Functions

by PatrickJMT

Here's a video by patrickJMT showing you how to take the derivative of inverse trigonometric functions.

Inverse trig derivatives aren't any harder than other derivatives but they can be a pain to memorize and remember. I for one did not remember any of these rules after all these years since I took Calc 1.

Also remember that inverse trig operations can be expressed in multiple ways. $$\sin^{-1}=a\sin=\arcsin$$

Here's the list of the derivatives. $$\begin{align} \frac{d}{dx}\sin^{-1}x&=\frac{1}{\sqrt{1-x^2}}\\ \frac{d}{dx}\cos^{-1}x&=-\frac{1}{\sqrt{1-x^2}}\\ \frac{d}{dx}\tan^{-1}x&=\frac{1}{1+x^2}\\ \frac{d}{dx}\csc^{-1}x&=-\frac{1}{x\sqrt{x^2-1}}\\ \frac{d}{dx}\sec^{-1}x&=\frac{1}{x\sqrt{x^2-1}}\\ \frac{d}{dx}\cot^{-1}x&=-\frac{1}{1+x^2}\\ \end{align}$$

The first example he solves is: $$\begin{align} f(x)&=(\sin^{-1}x)^2\\ f'(x)&=2(\sin^{-1}x)\cdot\color{red}{[\sin^{-1}x]'}\\ &=2(\sin^{-1}x)\frac{1}{\sqrt{1-x^2}}\\ &=\boxed{\frac{2(\sin^{-1}x)}{\sqrt{1-x^2}}} \end{align}$$

The next example is: $$\begin{align} g(t)&=\cos^{-1}\sqrt{2t-1}\\ g'(t)&=-\frac{1}{\sqrt{1-(\sqrt{2t-1})^2}}\cdot\color{red}{[\sqrt{2t-1}]'}\\ &=-\frac{1}{\sqrt{1-(\sqrt{2t-1})^2}}\cdot\frac12(2t-1)^{-\frac12}\cdot\color{red}{[2t-1]'}\\ &=-\frac{1}{\sqrt{1-(\sqrt{2t-1})^2}}\cdot\frac12(2t-1)^{-\frac12}\cdot2\\ &=-\frac{1}{\sqrt{1-(\sqrt{2t-1})^2}}\cdot\frac{1}{\sqrt{2t-1}}\\ &=-\frac{1}{\sqrt{1-(2t-1)}}\cdot\frac{1}{\sqrt{2t-1}}\\ &=\boxed{-\frac{1}{\sqrt{2-2t}}\cdot\frac{1}{\sqrt{2t-1}}}\\ \end{align}$$

The final example he solves is: $$\begin{align} y&=\tan^{-1}\bigg(\frac xa\bigg)+\ln\sqrt{\frac{x-a}{x+a}} \end{align}$$ In this problem the letter $a$ is just some arbitrary constant. Just treat it as you would any other number. Also note that we can simplify this problem by using some logarithm properties. $$\begin{align} y&=\tan^{-1}\bigg(\frac xa\bigg)+\frac12\ln\frac{x-a}{x+a}\\ &=\tan^{-1}\bigg(\frac xa\bigg)+\frac12\ln(x-a)-\frac12\ln(x+a)\\ y'&=\frac{1}{1+(\frac xa)^2}\cdot\color{red}{\bigg[\frac xa\bigg]'}+\frac12\frac{1}{x-a}-\frac12\frac{1}{x+a}\\ &=\frac{1}{1+(\frac xa)^2}\cdot\frac1a+\frac{\color{red}{x+a}}{2(x-a)\color{red}{(x+a)}}-\frac{\color{red}{x-a}}{2\color{red}{(x-a)}(x+a)}\\ &=\frac{1}{1+(\frac xa)^2}\cdot\frac1a+\frac{(x+a)-(x-a)}{2(x-a)(x+a)}\\ &=\frac{1}{1+(\frac xa)^2}\cdot\frac1a+\frac{2a}{2(x^2-a^2)}\\ &=\frac{1}{1+(\frac xa)^2}\cdot\frac1a+\frac{a}{x^2-a^2}\\ &=\frac{1}{a[1+(\frac xa)^2]}+\frac{a}{x^2-a^2}\\ &=\frac{1}{a+\frac{ax^2}{a^2}}+\frac{a}{x^2-a^2}\\ &=\frac{1}{a+\frac{x^2}{a}}+\frac{a}{x^2-a^2}\\ &=\frac{1}{\frac{a^2}{a}+\frac{x^2}{a}}+\frac{a}{x^2-a^2}\\ &=\frac{1}{\frac{a^2+x^2}{a}}+\frac{a}{x^2-a^2}\\ &=\frac{a}{x^2+a^2}+\frac{a}{x^2-a^2}\\ &=\frac{a\color{red}{(x^2-a^2)}}{(x^2+a^2)\color{red}{(x^2-a^2)}}+\frac{a\color{red}{(x^2+a^2)}}{\color{red}{(x^2+a^2)}(x^2-a^2)}\\ &=\frac{a(x^2-a^2)+a(x^2+a^2)}{(x^2+a^2)(x^2-a^2)}\\ &=\frac{a(2x^2)}{x^4-a^4}\\ &=\boxed{\frac{2ax^2}{x^4-a^4}} \end{align}$$

Here's another video where he goes over another example of inverse trig derivatives. The problem he looks at is: $$\begin{align} y&=x\cos^{-1}(\sqrt x)\\ y'&=\color{red}{[x]'}\cos^{-1}(\sqrt x)+x\color{red}{[\cos^{-1}(\sqrt x)]'}\\ &=\cos^{-1}(\sqrt x)+x\cdot-\frac{1}{\sqrt{1-(\sqrt x)^2}}\cdot\color{red}{[\sqrt x]'}\\ &=\cos^{-1}(\sqrt x)-x\cdot\frac{1}{\sqrt{1-x}}\cdot\frac{1}{2\sqrt x}\\ &=\boxed{\cos^{-1}(\sqrt x)-\frac{\sqrt{x}}{2\sqrt{1-x}}}\\ \end{align}$$

In this next video he looks at a slightly harder example. The problem is: $$y=\sec^{-1}\sqrt{1+x^2}$$ As always, it's good practice to rewrite the problem without the radical symbol. $$\begin{align} y&=\sec^{-1}(1+x^2)^{\frac12}\\ y'&=\frac{1}{(1+x^2)^{\frac12}\sqrt{[(1+x^2)^{\frac12}]^2-1}}\cdot\color{red}{[(1+x^2)^{\frac12}]'}\\ &=\frac{1}{(1+x^2)^{\frac12}\sqrt{1+x^2-1}}\cdot\frac12(1+x^2)^{-\frac12}\cdot\color{red}{[1+x^2]'}\\ &=\frac{1}{(1+x^2)^{\frac12}\sqrt{1+x^2-1}}\cdot\frac1{2(1+x^2)^{\frac12}}\cdot2x\\ &=\frac{1}{(1+x^2)^{\frac12}\sqrt{1+x^2-1}}\cdot\frac1{(1+x^2)^{\frac12}}\cdot x\\ &=\frac{x}{(1+x^2)^{\frac12}(1+x^2)^{\frac12}\sqrt{1+x^2-1}}\\ &=\frac{x}{(1+x^2)\sqrt{x^2}}\\ &=\boxed{\frac{x}{(1+x^2)|x|}} \end{align}$$

In this final video he goes over one more harder example: $$y=\sin^{-1}\bigg(\frac{\cos x}{1+\sin x}\bigg)$$ This isn't going to be much fun. $$\begin{align} y'&=\frac{1}{\sqrt{1-(\frac{\cos x}{1+\sin x})^2}}\cdot\color{red}{\bigg[\frac{\cos x}{1+\sin x}\bigg]'}\\ &=\frac{1}{\sqrt{1-(\frac{\cos x}{1+\sin x})^2}}\cdot\frac{\color{red}{[\cos x]'}(1+\sin x)-\cos x\color{red}{[1+\sin x]'}}{[1+\sin x]^2}\\ &=\frac{1}{\sqrt{1-(\frac{\cos x}{1+\sin x})^2}}\cdot\frac{-\sin x(1+\sin x)-\cos x\cos x}{(1+\sin x)^2}\\ &=\frac{1}{\sqrt{1-(\frac{\cos x}{1+\sin x})^2}}\cdot\frac{-\sin x-\sin^2 x-\cos^2 x}{(1+\sin x)^2}\\ &=\frac{1}{\sqrt{1-(\frac{\cos x}{1+\sin x})^2}}\cdot\frac{-\sin x-1}{(1+\sin x)^2}\\ &=\frac{1}{\sqrt{\frac{(1+\sin x)^2}{(1+\sin x)^2}-\frac{\cos^2 x}{(1+\sin x)^2}}}\cdot\frac{-\sin x-1}{(1+\sin x)^2}\\ &=\frac{1}{\sqrt{\frac{(1+\sin x)^2}{(1+\sin x)^2}-\frac{\cos^2 x}{(1+\sin x)^2}}}\cdot\frac{-\sin x-1}{(1+\sin x)^2}\\ &=\frac{1}{\sqrt{\frac{1+2\sin x+\sin^2x-\cos^2x}{(1+\sin x)^2}}}\cdot\frac{-\sin x-1}{(1+\sin x)^2}\\ &=\frac{1}{\frac{\sqrt{1+2\sin x+\sin^2x-\cos^2x}}{1+\sin x}}\cdot\frac{-\sin x-1}{(1+\sin x)^2}\\ &=\frac{1+\sin x}{\sqrt{1+2\sin x+\sin^2x-\cos^2x}}\cdot\frac{-\sin x-1}{(1+\sin x)^2}\\ &=\frac{1}{\sqrt{1+2\sin x+\sin^2x-\cos^2x}}\cdot\frac{-\sin x-1}{1+\sin x}\\ &=\frac{-1(\sin x +1)}{\sqrt{1+2\sin x+\sin^2x-\cos^2x}(1+\sin x)}\\ &=\frac{-1}{\sqrt{1+2\sin x+\sin^2x-\cos^2x}}\\ \end{align}$$ Recall that $\cos^2+\sin^2=1$ so if we rearrange that we get $\cos^2=1-\sin^2$. $$\begin{align} y'&=\frac{-1}{\sqrt{1+2\sin x+\sin^2x-\color{red}{(1-\sin^2x)}}}\\ &=\boxed{\frac{-1}{\sqrt{2\sin x+2\sin^2x}}}\\ \end{align}$$

(Video) Deriving the Derivative of Inverse Tangent

by PatrickJMT

Here's a good video by patrickJMT showing you how to derive the derivative of inverse tangent. This is helpful because it can be hard to remember all the derivative formulas for inverse trig functions. Furthermore, this is a good procedure to remember because you can use a similar method to derive many derivative formulas, like logarithms.

So the function we have is $$y=\tan^{-1}x$$ and we want to know what $y'$ or, in other words, $\frac{dy}{dx}$. We could just differentiate both sides but this doesn't help since we don't know the derivative for $\tan^{-1}x$. So let's rearrange our function by taking the tangent of both sides. $$\tan y=\tan(\tan^{-1}x)$$ Since a function and its inverse cancel, this now becomes: $$\tan y=x$$ Now we can use implicit differentiation and take the derivative of both sides. $$\begin{align} \frac{d}{dx}(\tan y)&=\frac{d}{dx}(x)\\ \sec^2y\cdot\color{red}{[y]'}&=1\\ \sec^2y\frac{dy}{dx}&=1\\ \frac{dy}{dx}&=\frac{1}{\sec^2y} \end{align}$$ So we want our derivative to be in terms of $x$ but we have a $y$ in there. We do have the expression: $\tan y=x$. So if we can transform $\sec^2y$ somehow into something with $\tan y$. Luckily, we have a nice trig identity to help us out. $$\begin{align} \cos^2y+\sin^2y&=1\\ \frac{\cos^2y}{\cos^2y}+\frac{\sin^2y}{\cos^2y}&=\frac{1}{\cos^2y}\\ 1+\tan^2y&=\sec^2y \end{align}$$ So we can substitute this into our derivative above. $$\begin{align} \frac{dy}{dx}&=\frac{1}{\sec^2y}\\ &=\frac{1}{\color{red}{1+\tan^2y}} \end{align}$$ And since we have our expression $\tan y=x$, we can now express our derivative in terms of $x$. $$\boxed{\frac{dy}{dx}=\frac{1}{1+x^2}}$$