Derivative of Inverse Function

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

Fact. Suppose that function $$${y}={f{{\left({x}\right)}}}$$$ has unique inverse and has finite derivative $$${f{'}}{\left({x}\right)}\ne{0}$$$ then derivative of 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 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, $$${\left({\operatorname{arcsin}{{\left({x}\right)}}}\right)}'=\frac{{1}}{{\sqrt{{{1}-{{x}}^{{2}}}}}}$$$.