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  • Please solve RD Sharma class 12 chapter Differentiation exercise 10.4 question 25 maths textbook solution

Please solve RD Sharma class 12 chapter Differentiation exercise 10.4 question 25 maths textbook solution

Answers (1)

Answer:

\frac{d y}{d x}=\frac{2 x^{3}+y-x^{2} y \cos x y}{x\left(x^{2} \cos x y+1+2 x y\right)}

Hint:

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

Given:

\sin (x y)+\frac{y}{x}=x^{2}-y^{2}

Solution:

\sin (x y)+\frac{y}{x}=x^{2}-y^{2}

Differentiate w.r.t x

\frac{d}{d x}(\sin x y)+\frac{d}{d x}\left(\frac{y}{x}\right)=\frac{d\left(x^{2}\right)}{d x}-\frac{d\left(y^{2}\right)}{d x}

\frac{d(\sin x y)}{d(x y)} \times \frac{d(x y)}{d x}+\frac{d}{d x}\left(\frac{y}{x}\right)=2 x-\frac{d y^{2}}{d y} \times \frac{d y}{d x}                        [Using chain rule and \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}]

\cos x y \times\left(x \frac{d y}{d x}+y \frac{d x}{d x}\right)+\frac{x \frac{d y}{d x}-y \frac{d x}{d x}}{x^{2}}=2 x-2 y \frac{d y}{d x}

x \cos x y \frac{d y}{d x}+y \cos x y+\frac{x \frac{d y}{d x}-y}{x^{2}}=2 x-2 y \frac{d y}{d x}

x \cos x y \frac{d y}{d x}+y \cos x y+\frac{1}{x} \frac{d y}{d x}-\frac{y}{x^{2}}=2 x-2 y \frac{d y}{d x}

x \cos x y \frac{d y}{d x}+\frac{1}{x} \frac{d y}{d x}+2 y \frac{d y}{d x}=2 x+\frac{y}{x^{2}}-y \cos x y

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

\frac{d y}{d x}\left(\frac{x^{2} \cos x y+1+2 x y}{x}\right)=\frac{2 x^{3}+y-x^{2} y \cos x y}{x^{2}}

\frac{d y}{d x}\left(x^{2} \cos x y+2 x y+1\right)=\frac{1}{x}\left(2 x^{3}+y-x^{2} y \cos x y\right)

\frac{d y}{d x}=\frac{2 x^{3}+y-x^{2} y \cos x y}{x\left(x^{2} \cos x y+1+2 x y\right)}

Hence proved

 

Posted by

infoexpert26

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