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  • need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.4 question 15

need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.4 question 15

Answers (1)

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Answer:

2 \frac{d y}{d x}+y^{3}=0

Hint:

Use product rule and \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}

Given:

x y^{2}=1

Solution:

Differentiate \left[x y^{2}=1\right]  w.r.t x

\frac{d\left(x y^{2}\right)}{d x}=\frac{d(1)}{d x}

x \frac{d y^{2}}{d x}+y^{2} \frac{d x}{d x}=0                \left[\because \frac{d(\operatorname{cons} \tan t)}{d x}=0\right]    [Product rule \frac{d(u v)}{d x}=u \cdot \frac{d v}{d x}+v \cdot \frac{d u}{d x}]

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

x(2 y) \times \frac{d y}{d x}+y^{2}=0                        \left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]

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

\begin{aligned} &\frac{d y}{d x}=-\frac{y^{2}}{2 x y}=\frac{-y}{2 x} \\ &2 \frac{d y}{d x}=-\frac{y}{x} \end{aligned}

2 \frac{d y}{d x}=-\frac{y}{\left(\frac{1}{y^{2}}\right)}                    \left[\begin{array}{c} x y^{2}=1 \\ x=\frac{1}{y^{2}} \end{array}\right]

2 \frac{d y}{d x}+y^{3}=0

Hence, if  x y^{2}=1  then 2 \frac{d y}{d x}+y^{3}=0  is the required answer

Hence proved

 

Posted by

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