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  • Please solve RD Sharma class 12 chapter Differentiation exercise 10.4 question 17 maths textbook solution

Please solve RD Sharma class 12 chapter Differentiation exercise 10.4 question 17 maths textbook solution

Answers (1)

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Answer:

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

Hint:

Use quotient rule and properties of logarithm

Given:

\log \sqrt{x^{2}+y^{2}}=\tan ^{-1}\left(\frac{y}{x}\right)

Solution:

\log \sqrt{x^{2}+y^{2}}=\tan ^{-1}\left(\frac{y}{x}\right)

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

\frac{1}{2} \log \left(x^{2}+y^{2}\right)=\tan ^{-1}\left(\frac{y}{x}\right) \quad\left[\because \log a^{m}=m \log a\right]

Differentiate this above equation w.r.t x

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

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

                                                                                             \left[\because \frac{d \log x}{d x}=\frac{1}{x}, \frac{d \tan ^{-1} x}{d x}=\frac{1}{1+x^{2}}, \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}\right]

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

\begin{aligned} &\text { - }\\ &\frac{1}{2\left(x^{2}+y^{2}\right)} \times\left(2 x+2 y \frac{d y}{d x}\right)=\frac{1}{\left(\frac{x^{2}+y^{2}}{x^{2}}\right)} \cdot\left(\frac{x}{x^{2}} \cdot \frac{d y}{d x}-\frac{y}{x^{2}}\right) \end{aligned}

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

 

\frac{x}{x^{2}+y^{2}}+\frac{y}{x^{2}+y^{2}} \cdot \frac{d y}{d x}=\left[\frac{x^{2}}{x^{2}+y^{2}} \times \frac{1}{x} \cdot \frac{d y}{d x}\right]-\frac{x^{2}}{x^{2}+y^{2}} \times \frac{y}{x^{2}}

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

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

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

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

Hence proved

 

Posted by

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