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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability2 Marks
Question
Differentiate the function $(5x)^{3\ \cos 2x}$ w.r.t to $x$.
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Answer

Let $y = (5x)^{3\ \cos\ 2x}$
Then, $\log y = \log (5x)^{3 \cos 2x}$
$\Rightarrow \log y=3 \cos 2 x \times \log 5 x$
Differentiating both sides with respect to $x$, we get
$\frac{1}{y} \frac{d y}{d x}$ = $3\left[\log 5 x \times \frac{d}{d x}(\cos 2 x)+\cos 2 x \times \frac{d}{d x}(\log 5 x)\right]$ ...[$\because$ $\frac{d}{d x}(u v)=u \times \frac{d v}{d x}+v \times \frac{d u}{d x}$]
$\Rightarrow$ $\frac{\mathrm{dy}}{\mathrm{dx}}=3 \mathrm{y}\left[\log 5 \mathrm{x}(-2 \sin 2 \mathrm{x}) \times \frac{\mathrm{d}}{\mathrm{dx}}(2 \mathrm{x})+\cos 2 \mathrm{x} \times \frac{1}{5 \mathrm{x}} \times \frac{\mathrm{d}}{\mathrm{dx}}(5 \mathrm{x})\right]$
$\Rightarrow$ $\frac{d y}{d x}=3 y\left[-2 \sin 2 x \log 5 x+\frac{\cos 2 x}{x}\right]$
$\Rightarrow$ $\frac{d y}{d x}=y\left[\frac{3 \cos 2 x}{x}-6 \sin 2 x \log 5 x\right]$
$\Rightarrow$ $\frac{d y}{d x}=(5 x)^{3} \cos 2 x\left[\frac{3 \cos 2 x}{x}-6 \sin 2 x \log 5 x\right]$
$\therefore$ $\frac{d y}{d x}=(5 x)^{3} \cos 2 x\left[\frac{3 \cos 2 x}{x}-6 \sin 2 x \log 5 x\right]$

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