How To Find The Derivative Of An Inverse Function?

If the one-to-one function f(x) is continuous and differentiable then its inverse function f^{-1}(x) is also continuous and differentiable.

Inverse Function Theorem:

If f(x) is a differentiable function with an inverse function f^{-1}(x) on an interval where f'(x) is nonzero, then \left(f^{-1}\right)'\left(x\right)=\frac{1}{f'\left(f^{-1}\left(x\right)\right)}.

So at point x=a, the derivative of the inverse function is,

\left(f^{-1}\right)'\left(a\right)=\frac{1}{f'\left(f^{-1}\left(a\right)\right)}               ……….(1)

Suppose the point (b, a) is on the graph of f(x),

So we can write, f(b)=a or we can write f^{-1}\left(a\right)=b\\

Substitute the value of f^{-1}\left(a\right) in equation 1.

\left(f^{-1}\right)'\left(a\right)=\frac{1}{f'\left(b\right)}\\

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Examples of the Derivative Of An Inverse Function:

Example 1: Let f(x)=x^3+e^x, Then \left(f^{-1}\right)'\left(1+e\right) equals?

Answer:

f(x)=x^3+e^x\\

Substitute x=1 and check.

f(1)=1^3+e^1\\

f(1)=1+e\\

So the point (b, a) is on the graph of f(x).

Where a=1+e and b=1.

 

f(x)=x^3+e^x\\

Take derivative.

f'(x)=\frac{d}{dx}\left(x^3+e^x\right)\\

f'(x)=3x^2+e^x\\

Substitute b=x=1

f'(1)=3(1)^2+e^1\\

f'(1)=3+e

 

Now use the Inverse Function Theorem:

\left(f^{-1}\right)'\left(a\right)=\frac{1}{f'\left(b\right)}\\

Substitute a=1+e and b=1

\left(f^{-1}\right)'\left(1+e\right)=\frac{1}{f'\left(1\right)}\\

\left(f^{-1}\right)'\left(1+e\right)=\frac{1}{3+e}\\

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Example 2: For the function f(x)=x^3+3\sin(x)+2\cos(x), find the value of \left(f^{-1}\right)'\left(2\right).

Answer:

f(x)=x^3+3\sin(x)+2\cos(x)\\

Substitute x=0 and check.

f(0)=0^3+3\sin(0)+2\cos(0)\\

f(0)=0+0+2\\

f(0)=2\\

So the point (b, a) is on the graph of f(x).

Where a=2 and b=0.

 

f(x)=x^3+3\sin(x)+2\cos(x)\\

Take derivative.

f'(x)=\frac{d}{dx}\left(x^3+3\sin \left(x\right)+2\cos \left(x\right)\right)\\

f'(x)=\frac{d}{dx}\left(x^3\right)+\frac{d}{dx}\left(3\sin \left(x\right)\right)+\frac{d}{dx}\left(2\cos \left(x\right)\right)\\

f'(x)=3x^2+3\cos \left(x\right)-2\sin \left(x\right)\\

Substitute b=x=0\\

f'(0)=3(0)^2+3\cos \left(0\right)-2\sin \left(0\right)\\

f'(0)=0+3-0\\

f'(0)=3\\

Now use the Inverse Function Theorem:

\left(f^{-1}\right)'\left(a\right)=\frac{1}{f'\left(b\right)}\\

Substitute a=2 and b=0

\left(f^{-1}\right)'\left(2\right)=\frac{1}{f'\left(0\right)}\\

\left(f^{-1}\right)'\left(2\right)=\frac{1}{3}\\