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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)

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

 

Posted by

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