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Explain solution RD Sharma class 12 chapter 10 Differentiation exercise Multiple choice question 12 maths

Answers (1)

Answer:

         -1

Hint:

        Differentiate the function w.r.t x

Given:

        \sqrt{x}+\sqrt{y}=1, \frac{d y}{d x} a t\left(\frac{1}{4}, \frac{1}{4}\right)

Solution:  

        \sqrt{x}+\sqrt{y}=1

   Differentiating w.r.t x 

        \begin{aligned} &\frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \frac{d y}{d x}=0 \\\\ &\frac{1}{2 \sqrt{y}} \frac{d y}{d x}=\frac{-1}{2 \sqrt{x}} \end{aligned}    

        \frac{d y}{d x}=-\frac{1}{2 \sqrt{x}} \times \frac{2 \sqrt{y}}{1}=\frac{-\sqrt{y}}{\sqrt{x}}

Now,

        \left[\frac{d y}{d x}\right]_{\left(\frac{1}{4} \cdot \frac{1}{4}\right)}=\frac{-\sqrt{\frac{1}{4}}}{\sqrt{\frac{1}{4}}}=-1

 

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