Limit Continuity And Differentiability Ques 20
- If $y=f\left(\frac{2 x-1}{x^2+1}\right)$ and $f^{\prime}(x)=\sin ^2 x$, then $\frac{d y}{d x}=\ldots \ldots$.
$(1982,2 \mathrm{M})$
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Answer:
Correct Answer: 20.$(\frac{-2\left(x^2-x-1\right)}{\left(x^2+1\right)^2} \cdot \sin ^2\left(\frac{2 x-1}{x^2+1}\right))$
Solution: Given, $\quad y=f\left(\frac{2 x-1}{x^2+1}\right)$
and $f^{\prime}(x)=\sin ^2 x$
$\therefore \quad \frac{d y}{d x}=f^{\prime}\left(\frac{2 x-1}{x^2+1}\right) \cdot \frac{d}{d x}\left(\frac{2 x-1}{x^2+1}\right)$
$=\sin ^2\left(\frac{2 x-1}{x^2+1}\right) \cdot\left\{\frac{\left(x^2+1\right) \cdot 2-(2 x-1)(2 x)}{\left(x^2+1\right)^2}\right\}$
$ \begin{aligned} & =\sin ^2\left(\frac{2 x-1}{x^2+1}\right) \cdot \frac{-2 x^2+2 x+2}{\left(x^2+1\right)^2} \\ & = \frac{-2\left(x^2-x-1\right)}{\left(x^2+1\right)^2} \sin ^2\left(\frac{2 x-1}{x^2+1}\right) \end{aligned} $
BETA
