# Derivative of Inverse Function

Now, let's use implicit differentiation to find derivatives of inverse functions.

Fact. Suppose that the function ${y}={f{{\left({x}\right)}}}$ has unique inverse and has finite derivative ${f{'}}{\left({x}\right)}\ne{0}$ then the derivative of the inverse function ${y}={{f}}^{{-{1}}}{\left({x}\right)}$ is ${\left({{f}}^{{-{1}}}{\left({x}\right)}\right)}'=\frac{{1}}{{{f{'}}{\left({{f}}^{{-{1}}}{\left({x}\right)}\right)}}}$.

Proof.

Suppose that we have a function ${y}={{f}}^{{-{1}}}{\left({x}\right)}$. This means that ${f{{\left({y}\right)}}}={x}$.

Differentiating this equality implicitly with respect to ${x}$ gives ${f{'}}{\left({y}\right)}{y}'={1}$ or ${y}'=\frac{{1}}{{{f{'}}{\left({y}\right)}}}$.

But ${y}={{f}}^{{-{1}}}{\left({x}\right)}$, so ${y}'=\frac{{1}}{{{f{'}}{\left({{f}}^{{-{1}}}{\left({x}\right)}\right)}}}$.

Thus, ${\left({{f}}^{{-{1}}}{\left({x}\right)}\right)}'=\frac{{1}}{{{f{'}}{\left({{f}}^{{-{1}}}{\left({x}\right)}\right)}}}$.

Now we can easily find derivative of inverse functions.

Example 1. Find derivative of logarithmic function ${y}={\log}_{{a}}{\left({x}\right)}$.

Logarithmic function is inverse of exponential, so here ${f{{\left({x}\right)}}}={{a}}^{{x}}$ and ${{f}}^{{-{1}}}{\left({x}\right)}={\log}_{{a}}{\left({x}\right)}$.

Since ${f{'}}{\left({x}\right)}={{a}}^{{x}}{\ln{{\left({a}\right)}}}$ then ${f{'}}{\left({{f}}^{{-{1}}}{\left({x}\right)}\right)}={f{'}}{\left({\log}_{{a}}{\left({x}\right)}\right)}={{a}}^{{{\log}_{{a}}{\left({x}\right)}}}{\ln{{\left({a}\right)}}}={x}{\ln{{\left({a}\right)}}}$.

So ${\left({{f}}^{{-{1}}}{\left({x}\right)}\right)}'=\frac{{1}}{{{f{'}}{\left({{f}}^{{-{1}}}{\left({x}\right)}\right)}}}=\frac{{1}}{{{x}{\ln{{\left({a}\right)}}}}}$.

Therefore, ${\left({\log}_{{a}}{\left({x}\right)}\right)}'=\frac{{1}}{{{x}{\ln{{\left({a}\right)}}}}}$.

Example 2. Find derivative of inverse sine function ${y}={\operatorname{arcsin}{{\left({x}\right)}}}$.

Inverse sine is inverse of sine function, so here ${f{{\left({x}\right)}}}={\sin{{\left({x}\right)}}}$ and ${{f}}^{{-{1}}}{\left({x}\right)}={\operatorname{arcsin}{{\left({x}\right)}}}$.

Since ${f{'}}{\left({x}\right)}={\cos{{\left({x}\right)}}}$ then ${f{'}}{\left({{f}}^{{-{1}}}{\left({x}\right)}\right)}={f{'}}{\left({\operatorname{arcsin}{{\left({x}\right)}}}\right)}={\cos{{\left({\operatorname{arcsin}{{\left({x}\right)}}}\right)}}}$.

So ${\left({{f}}^{{-{1}}}{\left({x}\right)}\right)}'=\frac{{1}}{{{f{'}}{\left({{f}}^{{-{1}}}{\left({x}\right)}\right)}}}=\frac{{1}}{{{\cos{{\left({\operatorname{arcsin}{{\left({x}\right)}}}\right)}}}}}$.

We can simplify ${\cos{{\left({\operatorname{arcsin}{{\left({x}\right)}}}\right)}}}$.

Using the main trigonometric identity, we have that ${{\cos}}^{{2}}{\left({\operatorname{arcsin}{{\left({x}\right)}}}\right)}+{{\sin}}^{{2}}{\left({\operatorname{arcsin}{{\left({x}\right)}}}\right)}={1}$.

From the properties of the inverse sine we know that ${\sin{{\left({\operatorname{arcsin}{{\left({x}\right)}}}\right)}}}={x}$. Also $-\frac{\pi}{{2}}\le{\operatorname{arcsin}{{\left({x}\right)}}}\le\frac{\pi}{{2}}$. That's why ${\cos{{\left({\operatorname{arcsin}{{\left({x}\right)}}}\right)}}}$ should be positive, so ${\cos{{\left({\operatorname{arcsin}{{\left({x}\right)}}}\right)}}}=\sqrt{{{1}-{{x}}^{{2}}}}$.

Thus, ${\left({\operatorname{arcsin}{{\left({x}\right)}}}\right)}'=\frac{{1}}{{\sqrt{{{1}-{{x}}^{{2}}}}}}$.