The derivative of an inverse function.

Definition. Let the function $y = f(x)$ be continuous and strictly monotonic in some neighborhood of the point $x_0$, and let the derivative $f'(x_0) \neq 0$ exist at this point. Then the inverse function at the point $y_0 = f(x_0)$ has a derivative, which can be found using the formula $\left(f^{-1}(y_0)\right)' = \frac{1}{f'(x_0)}$.

Examples.

Find the derivatives of the inverse functions $\left(f^{-1}(y)\right)'$.

1) $y = x + x^3$.

Solution.

$$\frac{dy}{dx}=1+3 x^2\Rightarrow\frac{dx}{dy}=\frac{1}{1+3x^2}.$$

Answer: $x'_y=\frac{1}{1+3x^2}.$

2) Find $\left(f^{-1}(0)\right)'$ and $\left(f^{-1}\left(\frac{6}{5}\right)\right)'$, where $y = x + \frac{1}{5}x^5$.

Solution:

If $y = 0$, then:

$0 = x + \frac{1}{5}x^5 \Rightarrow 0 = x\left(1 + \frac{1}{5}x^4\right) \Rightarrow$

$\begin{cases} x = 0 \ 1 + \frac{1}{5}x^4 = 0 \end{cases} \Rightarrow$

$\begin{cases} x = 0 \ x^4 = -5 \end{cases} \Rightarrow x = 0.$

If $y = \frac{6}{5}$, then $\frac{6}{5} = x + \frac{1}{5}x^5 \Rightarrow x = 1$. (The function has a unique root because it is strictly monotonic).

$y' = 1 + x^4 \Rightarrow x' = \frac{1}{1 + x^4}.$ Thus,

$$x'(0)=\frac{1}{1}=1; \,\, x'(6/5)=\frac{1}{1+1}=\frac{1}{2}.$$

Answer: $x'(0)=1; ,, x'\left(\frac{6}{5}\right)=\frac{1}{2}.$

3) $y=2x-\frac{\cos x}{2},,, y=-\frac{1}{2}.$

Solution.

$2x-\frac{\cos x}{2}=-\frac{1}{2},$ hence $x=0.$

$y'=2+\frac{\sin x}{2},$ therefore $x'=\frac{1}{2+\frac{\sin x}{2}}.$ Thus, $x'\left(-\frac{1}{2}\right)=\frac{1}{2}.$

Answer: $x'\left(-\frac{1}{2}\right)=\frac{1}{2}.$

4) $y=0.1x+e^{0.1x},,, y=1.$

Solution.

$0.1x+e^{0.1x}=1, $ hence $x=0.$

$y'=0.1+0.1e^{0.1x},$ therefore $x'=\frac{1}{0.1+0.1e^{0.1x}}.$

Thus, $x'(1)=\frac{1}{\frac{2}{10}}=5.$

Answer: $x'(1)=5.$