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  • Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.4 question 14

Provide solution for RD Sharma maths class 12 chapter Differentiation exercise  10.4 question 14

Answers (1)

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Answer:

\frac{d y}{d x}+y^{2}=0

Hint:

Use product rule

Given:

x y=1

Solution:

Differentiate x y=1  w.r.t x

\frac{d(x y)}{d x}=\frac{d(1)}{d x}

x \frac{d y}{d x}+y \frac{d x}{d x}=0                            \left[\begin{array}{l} \because \frac{d(\operatorname{cons} \tan t)}{d x}=0 \\ \frac{d(u \cdot v)}{d x}=u \cdot \frac{d v}{d x}+v \cdot \frac{d u}{d x} \end{array}\right]

x \frac{d y}{d x}+y=0

\begin{aligned} &x \frac{d y}{d x}=-y \\ &\frac{d y}{d x}=-\frac{y}{x} \end{aligned}

\frac{d y}{d x}=-\frac{y}{\left(\frac{1}{y}\right)} \; \; \; \; \; \; \; \quad\left[\begin{array}{c} x y=1 \\ x=\frac{1}{y} \end{array}\right]

\frac{d y}{d x}=-y^{2} \; \; \; \; \; \; \; \; \; \; \quad\left[\frac{d y}{d x}+y^{2}=0\right]

Hence, if   x y=1

Then \frac{d y}{d x}+y^{2}=0  hence proved

 

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