Differentiate the function $(5 x)^{3 cos 2 x}$ w.r.t. $x$ .

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Solution

Let $y = (5 x)^{3 cos 2 x}$

Taking $log$ on both sides, we get

$log y = 3 cos 2 x . log 5 x$

$(∵ log (a^{n}) = n . log a)$

Differentiating both sides w.r.t. $x$ , we get

$\frac{d (log y)}{d x} = \frac{d (3 cos 2 x . log 5 x)}{d x}$

$\frac{d (log y)}{d y} (\frac{d y}{d x}) = \frac{d (3 cos 2 x . log 5 x)}{d x}$

$\frac{1}{y} . \frac{d y}{d x} = \frac{d (3 cos 2 x . log 5 x)}{d x}$

Using Product Rule : $(u v)^{'} = u^{'} v + v^{'} u$

$\frac{1}{y} . \frac{d y}{d x} = \frac{d (3 cos 2 x)}{d x} . log 5 x + \frac{d (log 5 x)}{d x} .3 cos 2 x$

Using chain rule : $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$

$\frac{1}{y} . \frac{d y}{d x} = 3. \frac{d (cos 2 x)}{d x} . log 5 x + \frac{1}{5 x} . \frac{d (5 x)}{d x} .3 cos 2 x$

$\frac{1}{y} . \frac{d y}{d x} = 3. (- sin 2 x) \times 2 \times log 5 x + \frac{1}{5 x} \times 5 \times 3 cos 2 x$

$\frac{1}{y} . \frac{d y}{d x} = - 6 sin 2 x . log 5 x + \frac{3 cos 2 x}{x}$

$\frac{d y}{d x} = y (- 6 sin 2 x . log 5 x + \frac{3 cos 2 x}{x})$

Substituing $y = (5 x)^{3 cos 2 x}$ ,

$∴ \frac{d y}{d x} = (5 x)^{3 cos 2 x} (\frac{3 cos 2 x}{x} - 6 sin 2 x . log 5 x)$

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