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

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

Answers (1)

best_answer

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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