What is Written 1

Calculus Level 4

Let $f(x)$ be a real valued bijective function satisfying $f'(x) = \sin^2 (\sin (x+1))$ and $f(0) = 3$ . Find the value of $\left( f^{-1}\right)' (3)$ .

Clarifications :

$f'(x)$ denotes the first derivative of $f(x)$ with respect to $x$ .
$f^{-1} (x)$ denotes the inverse function of $f(x)$ .

\cos ^{ 2 }{ \left( \sin { 1 } \right) }

2\cos { \left( \sin { 4 } \right) } \sin { 4 }

\frac { 1 }{ \cos ^{ 2 }{ \left( \sin { 1 } \right) } } \quad

\frac { 1 }{ 2\cos { \left( \sin { 4 } \right) } \sin { 4 } }

\sin ^{ 2 }{ \sin { 1 } }

\cos ^{ 2 }{ \left( \cos { 4 } \right) }

\frac { 1 }{ \sin ^{ 2 }{ \sin { 1 } } }

\frac { 1 }{ \cos ^{ 2 }{ \left( \cos { 4 } \right) } } \quad

1 solution

Sahil Bansal
May 20, 2016

Since $f$ is a bijective function, hence $f(f^{-1}(x))=x$ .

And, we need to find derivative of $f^{-1}(x)$ to get the value of $(f^{-1})^{'}(3)$ , so we differentiate the above equation using chain rule:

$f'(f^{-1}(x)).(f^{-1})^{'}(x)=1$

$\Rightarrow (f^{-1})^{'}(x)=\frac{1}{f'(f^{-1}(x))}$

Putting $x=3$ in the above equation:

$\Rightarrow (f^{-1})^{'}(3)=\frac{1}{f'(f^{-1}(3))}$

Now, since $f(0)=3 \Rightarrow f^{-1}(3)=0$

$\Rightarrow (f^{-1})^{'}(3)=\frac{1}{f'(0)}$

$\Rightarrow (f^{-1})^{'}(3)=\frac{1}{\sin ^{2}\sin 1}$

What is Written 1

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