Inverse Functions

Proof: the derivative of y=sin^-1(x) using implicit differentiation.

We can rewrite y=sin^-1(x) into x=sin(y) because that is the definition of an inverse function (recall that we just switch the x and the y to get the inverse function). You can also think of it as taking the sine of both sides.

Now we can implicitly differentiate both sides since we have rewritten our function in terms of x.

$\frac{d}{dx}x=\frac{d}{dx}\sin y$

$1=\cos y\cdot \frac{dy}{dx}$

$\frac{dy}{dx}=\frac{1}{\cos y}$, but we don’t like y-terms in our final derivative. Recall that we can find cos(y) in terms of sin(y) via our trig identities,

${\cos }^{2}y+{\sin }^{2}y=1$

$\cos y=\sqrt{1-{\sin }^{2}y}$

We can now substitute this into the purple derivative formula. Remember that sin y = x from the original definition of an inverse function. Finally, we get

$\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}{\mathbf{sin}}^{\mathbf{-}\mathbf{1}}\mathit{x}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{-}{\mathit{x}}^{\mathbf{2}}}}$

Proof: the derivative of y=tan^-1(x) using implicit differentiation.

Same strategy as before. First we rearrange the equation to get x=tan(y), then differentiate both sides with respect to x.

$\frac{d}{dx}x=\frac{d}{dx}(\tan y)$

$1={\sec }^{2}y\cdot \frac{dy}{dx}$

$\frac{dy}{dx}=\frac{1}{{\sec }^{2}y}$

Again, we refer to our trig identities to see that sec²y=1+tan²y. Since tan(y)=x, we can substitute these two values into the derivative function so that we get

$\frac{dy}{dx}{\tan }^{-1}x=\frac{1}{1+x^2}$

1. In order to use the formula, we must calculate the derivative of the original function as well as the inverse of the original function. As a reminder, the formula is:

$\frac{\mathit{d}}{\mathit{d}\mathit{x}}{\mathit{f}}^{\mathbf{-}\mathbf{1}}\mathbf{(}\mathit{x}\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{{\mathit{f}}^{\mathbf{\prime }}\mathbf{(}{\mathit{f}}^{\mathbf{-}\mathbf{1}}\mathbf{(}\mathit{x}\mathbf{)}\mathbf{)}}$

We’ll start by creating a table and filling it in as we go.

$\begin{array}{rlrl}f(x)& =x^2+2x+1& \frac{d}{dx}f(x)& =2x+2\\ f^{-1}(x)& =?& \frac{d}{dx}{f}^{-1}(x)& =?\end{array}$

To calculate $y=f^{-1}(x)$, we simply switch the x and y in the original equation, which gives us $x=y^2+2y+1$, where $y=f^{-1}(x)$

$x=(y+1{)}^2$

$\sqrt{x}=y+1$

$f^{-1}(x)=\sqrt{x}-1$

For the next step, we simply plug this into the formula, but it’s tricky, so watch carefully!

$\frac{d}{dx}{f}^{-1}(x)=\frac{1}{f^{\prime }(f^{-1}(x))}=\frac{1}{2(\sqrt{x}-1)+2}$

Simplifying this gives us $\frac{\mathit{d}}{\mathit{d}\mathit{x}}{\mathit{f}}^{\mathbf{-}\mathbf{1}}\mathbf{(}\mathit{x}\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\sqrt{\mathit{x}}}$

2. Next, we try without the formula. We know that $f^{-1}(x)=\sqrt{x}-1$ from part a). Now we just take the derivative of this normally. The derivative of the constant (–1) is 0, and the derivative of $\sqrt{x}$ is $\frac{1}{2\sqrt{x}}$, therefore $\frac{\mathit{d}}{\mathit{d}\mathit{x}}{\mathit{f}}^{\mathbf{-}\mathbf{1}}\mathbf{(}\mathit{x}\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\sqrt{\mathit{x}}}$. We get the same answer!

Using the color coding scheme again:

$\text{Magenta}$ - Quotient Rule​
$\text{Green}$ - Chain Rule​
$\text{Blue}$ - Product Rule

Let’s take the first derivative:

$y^{\prime }=\frac{1}{1+{\sin }^{2}x}\cdot \cos x=\frac{\cos x}{1+{\sin }^{2}x}$

$y^{\prime \prime }=\frac{(1+{\sin }^{2}x)\cdot \frac{d}{dx}(\cos x)-\cos x\cdot \frac{d}{dx}(1+{\sin }^{2}x)}{(1+{\sin }^{2}x{)}^2}$

$y^{\prime \prime }=\frac{-\sin x(1+{\sin }^{2}x)-\cos x\cdot 2\sin x\cos x}{(1+{\sin }^{2}x{)}^2}$

At this point we can just go ahead and plug in $x=\frac{\pi }{2}$ into the derivative to get the answer (we do not need to simplify since we are looking for a numerical answer). We know that sin(π/2)=1 and cos(π/2)=0 from our unit circle. That’s all we need to solve the rest of this problem.

$\mathbf{a}\mathbf{t}\frac{\mathit{\pi }}{\mathbf{2}}\mathbf{,}{\mathit{y}}^{\mathbf{\prime }\mathbf{\prime }}\mathbf{=}\frac{\mathbf{-}\mathbf{1}\mathbf{\cdot }\mathbf{(}\mathbf{1}\mathbf{+}{\mathbf{1}}^{\mathbf{2}}\mathbf{)}\mathbf{-}\mathbf{0}\mathbf{\cdot }\mathbf{2}\mathbf{\cdot }\mathbf{1}\mathbf{\cdot }\mathbf{0}}{\mathbf{(}\mathbf{1}\mathbf{+}{\mathbf{1}}^{\mathbf{2}}{\mathbf{)}}^{\mathbf{2}}}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}$