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  • Need solution for RD Sharma maths class 12 chapter 10 Differentiation exercise Very short answers question 11

Need solution for RD Sharma maths class 12 chapter 10 Differentiation exercise Very short answers question 11

Answers (1)

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Answer:

The answer of the given question will be 2

Given:

\text { Iff }(0)=f(1)=0, f^{\prime}(1)=2 \text { and } y=f\left(e^{x}\right) e^{f(x)}  write the value of \frac{d y}{d x} \text { for } x=0

Hint:

\frac{d}{d x} u v=u \cdot \frac{d}{d x} \cdot v+. v \frac{d}{d x} u

Solution:  

\frac{d y}{d x}=\left[f\left(e^{x}\right) e^{f(x)}\right]=\frac{d}{d x}\left[f\left(e^{x}\right) \cdot e^{f(x)}+f\left(e^{x}\right) \cdot \frac{d}{d x} e^{f(x)}\right]

      =\frac{d}{d x} f^{\prime}\left(e^{x}\right) \cdot \frac{d}{d x} e^{x} \cdot e^{f(x)}+f\left(e^{x}\right) \cdot e^{f(x)} \cdot \frac{d}{d x}[f(x)]

\begin{aligned} &\Rightarrow \frac{d y}{d x}=f^{\prime}\left(e^{x}\right) \cdot e^{x} \cdot e^{f(x)}+f\left(e^{x}\right) \cdot e^{f(x)} \cdot f^{\prime}(x) \\\\ &\Rightarrow \frac{d y}{d x}=f^{\prime}\left(e^{0}\right) \cdot e^{0} \cdot e^{f(0)}+f\left(e^{0}\right) \cdot e^{f(0)} \cdot f^{\prime}(0) \end{aligned}

\begin{aligned} &=f^{\prime}(1) \cdot 1 \cdot e^{f(0)}+f(1) \cdot e^{f(0)} \cdot f^{\prime}(0) \\\\ &=2 . e^{0}+0 . e^{0} \cdot 2 \\\\ &=2.1+0 \\\\ &=2 \end{aligned}

So the answer of \frac{d y}{d x}=2

 

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