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If for a continuous function \mathrm{f, f(0)=f(1)=0, f^{\prime}(1)=2 ~and ~y(x)=f\left(e^x\right) e^{f(x)}, ~then ~y^{\prime}(0)} is equal to.

Option: 1

1


Option: 2

2


Option: 3

0


Option: 4

None of these.


Answers (1)

best_answer

We have, \mathrm{y(x)=f\left(e^x\right) e^{f(x)}}
\mathrm{ \therefore y^{\prime}(x) =f^{\prime}\left(e^x\right) \cdot e^x \cdot e^{f(x)}+f\left(e^x\right) e^{f(x)} f^{\prime}(x) }
\mathrm{ \Rightarrow y^{\prime}(0) =f^{\prime}(1) e^{f(0)}+f(1) e^{f(0)} f^{\prime}(0) }
\mathrm{ =2 }          \mathrm{ \quad[\because f(0)=f(1)=0, f(1)=2]}

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