Limit Continuity And Differentiability Ques 56
- For $x>1$, if $(2 x)^{2 y}=4 e^{2 x-2 y}$, then $\left(1+\log _e 2 x\right)^2 \frac{d y}{d x}$ is equal to
(2019 Main, 12 Jan I)
(a) $\frac{x \log _e 2 x+\log _e 2}{x}$
(b) $\frac{x \log _e 2 x-\log _e 2}{x}$
(c) $x \log _e 2 x$
(d) $\log _e 2 x$
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Answer:
Correct Answer: 56.(b)
Solution: (b) Given equation is
$ (2 x)^{2 y}=4 \cdot e^{2 x-2 y} $
On applying $ ’ \log _e’ $ both sides, we get
$\log _e(2 x)^{2 y}=\log _e 4+\log _e e^{2 x-2 y} $
$2 y \log _e(2 x)=\log _e(2)^2+(2 x-2 y) $
${\left[\because \log _e n^m=m \log _e n \text { and } \log _e e^{f(x)}=f(x)\right]} $
$\Rightarrow \quad \left(2 \log _e(2 x)+2\right) y=2 x+2 \log _e(2) $
$\Rightarrow \quad y=\frac{x+\log _e 2}{1+\log _e(2 x)}$
On differentiating ’ $y$ ’ w.r.t. ’ $x$ ‘, we get
$ \begin{aligned} \frac{d y}{d x} & =\frac{\left(1+\log _e(2 x)\right) 1-\left(x+\log _e 2\right) \frac{2}{2 x}}{\left(1+\log _e(2 x)\right)^2} \\ & =\frac{1+\log _e(2 x)-1-\frac{1}{x} \log _e 2}{\left(1+\log _e(2 x)\right)^2} \end{aligned} $
So, $\left(1+\log _e(2 x)\right)^2 \frac{d y}{d x}=\left(\frac{x \log _e(2 x)-\log _e 2}{x}\right)$
BETA
