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Explain solution RD Sharma class 12 chapter 10 Differentiation exercise Very short answers question 24 maths

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Answer:

The value of f^{\prime}(1) \text { is } 0

Hint:

Using the chain rule of differentiation.

Given:

f(x)=\log \left\{\frac{u(x)}{v(x)}\right\}, u(1)=v(1) \text { and } u^{\prime}(1)=v^{\prime}(1)=2
Solution:  

Using the chain rule of differentiation

f^{\prime}(x)=\frac{1}{\frac{u(x)}{v(x)}} \cdot \frac{v(x) \cdot u^{\prime}(x)-v^{\prime}(x) \cdot u(x)}{(v(x))^{2}}

f^{\prime}(x)=\frac{v(x) \cdot u^{\prime}(x)-v^{\prime}(x) \cdot u(x)}{u(x) \cdot v(x)}

Putting x=1

f^{\prime}(1)=\frac{v(1) \cdot u^{\prime}(1)-v^{\prime}(1) \cdot u(1)}{u(1) \cdot v(1)}

           =\frac{2 v(1) \cdot-2 u(1)}{u(1) \cdot v(1)}

Since,

\begin{aligned} &u(1)=v(1) \\\\ &2 v(1) \cdot-2 v(1)=0 \\\\ &\text { i.e, } f^{\prime}(1)=0 \end{aligned}

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