# 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(x) has unique inverse and has finite derivative f'(x)!=0 then derivative of inverse function y=f^(-1)(x) is (f^(-1)(x))'=1/(f'(f^(-1)(x))).

Proof.

Suppose that we have function y=f^(-1)(x). This means that f(y)=x.

Differentiating this equality implicitly with respect to x gives f'(y)y'=1 or y'=1/(f'(y)).

But y=f^(-1)(x), so y'=1/(f'(f^(-1)(x))).

Thus, (f^(-1)(x))'=1/(f'(f^(-1)(x))).

Now we can easily find derivative of inverse functions.

Example 1. Find derivative of logarithmic function y=log_a(x).

Logarithmic function is inverse of exponential, so here f(x)=a^x and f^(-1)(x)=log_a(x).

Since f'(x)=a^xln(a) then f'(f^(-1)(x))=f'(log_a(x))=a^(log_a(x))ln(a)=xln(a).

So (f^(-1)(x))'=1/(f'(f^(-1)(x)))=1/(xln(a)).

Therefore, (log_a(x))'=1/(xln(a)).

Example 2. Find derivative of inverse sine function y=arcsin(x).

Inverse sine is inverse of sine function, so here f(x)=sin(x) and f^(-1)(x)=arcsin(x).

Since f'(x)=cos(x) then f'(f^(-1)(x))=f'(arcsin(x))=cos(arcsin(x)).

So (f^(-1)(x))'=1/(f'(f^(-1)(x)))=1/(cos(arcsin(x))).

We can simplify cos(arcsin(x)).

Using main trigonometric identity we have that cos^2(arcsin(x))+sin^2(arcsin(x))=1.

From properties of inverse sine we know that sin(arcsin(x))=x. Also -pi/2<=arcsin(x)<=pi/2. That's why cos(arcsin(x)) should be positive, so cos(arcsin(x))=sqrt(1-x^2).

Thus, (arcsin(x))'=1/(sqrt(1-x^2)).