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  • Please solve RD Sharma class 12 chapter Differentiation exercise 10.2 question 32 maths textbook solution

Please solve RD Sharma class 12 chapter Differentiation exercise 10.2 question 32 maths textbook solution

Answers (1)

Answer:  \frac{-2\left(x^{2}-1\right)}{x^{4}+x^{2}+1}

Hint: You must know about the rules of solving derivative of logarithm functions.

Given: \log \left(\frac{x^{2}+x+1}{x^{2}-x+1}\right)

Solution:

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

 

Differentiate with respect to x,

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

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

 

Now apply quotient rule,  \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}

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

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

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

\frac{d y}{d x}=\frac{-4 x^{2}+2 x^{2}+2}{\left(x^{2}+x+1\right)\left(x^{2}-x+1\right)}

\frac{d y}{d x}=\frac{-4 x^{2}+2 x^{2}+2}{\left(x^{2}+1\right)^{2}-(x)^{2}}

\begin{aligned} &\frac{d y}{d x}=\frac{-2\left(x^{2}-1\right)}{x^{4}+1+2 x^{2}-x^{2}} \\ &\frac{d y}{d x}=\frac{-2\left(x^{2}-1\right)}{x^{4}+x^{2}+1} \end{aligned}

 

 

Posted by

infoexpert26

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