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 $f^{\prime }(x_0)\ne 0$ 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 ${(f^{-1}(y_0))}^{\prime }=\frac{1}{f^{\prime }(x_0)}$.

Examples.

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

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

Solution.

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

Answer: $x_y^{\prime }=\frac{1}{1+3x^2}.$

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

Solution:

If $y=0$, then:

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

$\{\begin{array}{l}x=01+\frac{1}{5}{x}^4=0\end{array}\Rightarrow$

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

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).

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

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

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

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

Solution.

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

$y^{\prime }=2+\frac{\sin x}{2},$ therefore $x^{\prime }=\frac{1}{2+\frac{\sin x}{2}}.$ Thus, $x^{\prime }(-\frac{1}{2})=\frac{1}{2}.$

Answer: $x^{\prime }(-\frac{1}{2})=\frac{1}{2}.$

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

Solution.

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

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

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

Answer: $x^{\prime }(1)=5.$

Tags: calculus, derivative, derivative table, inverse function, mathematical analysis