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Derivative of Inverse Function

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

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

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

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

Thus, (f−1(x))′=1f′(f−1(x)){\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(x){y}={\log}_{{a}}{\left({x}\right)}.

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

Since f′(x)=axln⁡(a){f{'}}{\left({x}\right)}={{a}}^{{x}}{\ln{{\left({a}\right)}}} then f′(f−1(x))=f′(log⁡a(x))=alog⁡a(x)ln⁡(a)=xln⁡(a){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 (f−1(x))′=1f′(f−1(x))=1xln⁡(a){\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, (log⁡a(x))′=1xln⁡(a){\left({\log}_{{a}}{\left({x}\right)}\right)}'=\frac{{1}}{{{x}{\ln{{\left({a}\right)}}}}}.

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

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

Since f′(x)=cos⁡(x){f{'}}{\left({x}\right)}={\cos{{\left({x}\right)}}} then f′(f−1(x))=f′(arcsin⁡(x))=cos⁡(arcsin⁡(x)){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 (f−1(x))′=1f′(f−1(x))=1cos⁡(arcsin⁡(x)){\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⁡(arcsin⁡(x)){\cos{{\left({\operatorname{arcsin}{{\left({x}\right)}}}\right)}}}.

Using the main trigonometric identity, we have that cos⁡2(arcsin⁡(x))+sin⁡2(arcsin⁡(x))=1{{\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⁡(arcsin⁡(x))=x{\sin{{\left({\operatorname{arcsin}{{\left({x}\right)}}}\right)}}}={x}. Also −π2≤arcsin⁡(x)≤π2-\frac{\pi}{{2}}\le{\operatorname{arcsin}{{\left({x}\right)}}}\le\frac{\pi}{{2}}. That's why cos⁡(arcsin⁡(x)){\cos{{\left({\operatorname{arcsin}{{\left({x}\right)}}}\right)}}} should be positive, so cos⁡(arcsin⁡(x))=1−x2{\cos{{\left({\operatorname{arcsin}{{\left({x}\right)}}}\right)}}}=\sqrt{{{1}-{{x}}^{{2}}}}.

Thus, (arcsin⁡(x))′=11−x2{\left({\operatorname{arcsin}{{\left({x}\right)}}}\right)}'=\frac{{1}}{{\sqrt{{{1}-{{x}}^{{2}}}}}}.