Inverse trig derivatives

What you’ll learn

• Inverse trig derivatives: Memorize the explicit derivative formulas.
• Chain rule application: Apply these formulas when the inside function is more than just $x$

Inverse trigonometric functions

Regular trigonometric functions take an angle $\theta$ and output a ratio.

Inverse trigonometric functions do the reverse: inputting a trigonometric ratio gives the angle that produces it (within the function’s restricted range).

$$\sin \left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}⟺\arcsin \left(\frac{\sqrt{2}}{2}\right)=\frac{\pi }{4}$$

Sidenote

Inverse notation

$\arcsin (x)$ is identical to ${\sin }^{-1}(x)$.

Make sure not to confuse this with the reciprocal:

$${\sin }^{-1}(x)\ne \frac{1}{\sin (x)}$$

The derivative formulas

While there are six inverse trig functions, the AP curriculum only tests the first three of them.

$f(x)$	$f^{\prime }(x)$
${\sin }^{-1}(x)$	$\frac{1}{\sqrt{1-x^2}}$
${\cos }^{-1}(x)$	$-\frac{1}{\sqrt{1-x^2}}$
${\tan }^{-1}(x)$	$\frac{1}{1+x^2}$
${\cot }^{-1}(x)$	$-\frac{1}{1+x^2}$
${\sec }^{-1}(x)$	$\frac{1}{|x|\sqrt{x^2-1}}$
${\csc }^{-1}(x)$	$-\frac{1}{|x|\sqrt{x^2-1}}$

Notice that every other derivative is just the negative of the one above it.

Optional reading: Deriving the formulas

If you ever forget the formula on exam day, you can actually build it yourself using implicit differentiation and a right triangle.

For example, to find the derivative of $y=\arcsin (x)$:

1. Rewrite it: Rearrange it into a normal trig function.

$$y=\arcsin (x)⟺x=\sin (y)$$

2. Differentiate implicitly: Take the derivative of both sides with respect to $x$:

$$\begin{array}{rl}\frac{d}{dx}(\sin y)& =\frac{d}{dx}(x)\\ \cos (y)\cdot \frac{dy}{dx}& =1\end{array}$$

3. Isolate $\frac{dy}{dx}$:

$$\frac{dy}{dx}=\frac{1}{\cos (y)}$$

4. Substitute back:

Use a right triangle where $\sin (y)=\frac{x}{1}$ (opposite over hypotenuse).

By the Pythagorean theorem, the adjacent side is $\sqrt{1-x^2}$. This is depicted in the triangle below (angle $\theta =y$).

From the triangle, $\cos (y)=\frac{\sqrt{1-x^2}}{1}$. Substituting this back in gives the derivative of $y=\arcsin x$:

$$\begin{array}{rl}\frac{dy}{dx}& =\frac{1}{\cos (y)}=\frac{1}{\sqrt{1-x^2}}\end{array}$$

Applying the chain rule

If the argument inside the function is anything other than a simple $x$, apply the chain rule:

• Replace every $x$ in the standard derivative formula with the new argument.
• Multiply by the derivative of that argument (this is the chain rule).

Find the derivative of

$$y=\arccos (e^x)$$

Solution

(spoiler)

The derivative of the basic function $y=\arccos (x)$ is

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

To differentiate $y=\arccos (e^x)$, replace $x$ in the formula with the argument $e^x$ and multiply by the derivative of $e^x$:

$$\begin{array}{rl}{y}^{\prime }& =-\frac{1}{\sqrt{1-(e^x{)}^2}}\cdot e^x\\ & =-\frac{e^x}{\sqrt{1-e^{2x}}}\end{array}$$

Simplifying a composite function

Find the derivative of

$$f(x)=\tan ({\cos }^{-1}(2x))$$

Solution

(spoiler)

Differentiate using the chain rule. Recall that $\frac{d}{dx}[\tan (u)]={\sec }^2(u)\cdot u^{\prime }$.

$$\begin{array}{rl}{f}^{\prime }(x)& ={\sec }^2({\cos }^{-1}(2x))\cdot \left(-\frac{2}{\sqrt{1-(2x{)}^2}}\right)\end{array}$$

Next, simplify the trigonometric expression to an entirely algebraic function of $x$.

Use a triangle with $\cos (\theta )=\frac{2x}{1}$, as depicted below.

Since $\sec (\theta )=\frac{1}{2x}$, then ${\sec }^2(\theta )=\frac{1}{4x^2}$. Therefore,

$$\begin{array}{rl}{f}^{\prime }(x)& =\frac{1}{4x^2}\cdot \frac{-2}{\sqrt{1-4x^2}}\\ & =-\frac{1}{2x^2\sqrt{1-4x^2}}\end{array}$$

AP-style problem

Find the equation of the tangent line to $y={\sin }^{-1}x$ at $x=\frac{1}{2}$.

Solution

(spoiler)

1. Find the point of tangency:

$$y=\arcsin (\frac{1}{2})=\frac{\pi }{6}$$

The point is $(\frac{1}{2},\frac{\pi }{3})$.

2. Find the slope $m$:

Find the derivative and evaluate it at $x=\frac{1}{2}$:

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

$$m=\frac{1}{\sqrt{1-{\left(\frac{1}{2}\right)}^2}}=\frac{1}{\sqrt{\frac{3}{4}}}=\frac{2}{\sqrt{3}}$$

3. Write the equation:

In point-slope form,

$$\boxed{y-\frac{\pi }{6}=\frac{2}{\sqrt{3}}\left(x-\frac{1}{2}\right)}$$