Let $$f$$ be a real-valued function defined on the interval (-1, 1) such that $${e^{ - x}}f\left( x \right) = 2 + \int\limits_0^x {\sqrt {{t^4} + 1} } \,dt,$$      for all $$x \in \left( { - 1,\,1} \right),$$   and let $${f^{ - 1}}$$  be the inverse function of $$f.$$  Then $$\left( {{f^{ - 1}}} \right)'\left( 2 \right)$$     is equal to-

Solution :
$$\eqalign{ & {e^{ - x}}f\left( x \right) = 2 + \int_0^x {\sqrt {1 + {t^4}} } \,dt\,\forall \,x \in \left( { - 1,\,1} \right) \cr & {\text{At }}x = 0,\,\,f\left( 0 \right) = 2 \cr} $$
Now on differentiating, we get
$$\eqalign{ & - {e^{ - x}}f\left( x \right) + {e^{ - x}}f'\left( x \right) = 0\sqrt {1 + {x^4}} \cr & \Rightarrow - f\left( 0 \right) + f'\left( 0 \right) = 1 \cr & \Rightarrow f'\left( 0 \right) = 3 \cr & {\text{Now }}{f^{ - 1}}\left( {f\left( x \right)} \right) = x \cr & \Rightarrow \left[ {\left( {{f^{ - 1}}} \right)'\left( {f\left( x \right)} \right)} \right]f'\left( x \right) = 1 \cr & \Rightarrow \left( {{f^{ - 1}}} \right)'\left( {f\left( 0 \right)} \right)f'\left( 0 \right) = 1 \cr & \Rightarrow \left( {{f^{ - 1}}} \right)'\left( 2 \right) = \frac{1}{3} \cr} $$