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))`.