Question

Differentiate w.r.t. $x$ the function: $(5x)^{3cos2x}$

Answer 1

The correct answer is $(5x)^{3cos2x}[\frac{3cos\,2x}{x}-6sin\,2x\,log\,5x]$
Let $y=(5x)^{3cos2x}$
Taking logarithm on both the sides, we obtain
$logy=3cos\,2x\,log\,5x$
Differentiating both sides with respect to $x$, we obtain
$\frac{1}{y}\frac{dy}{dx}=3[log5x.\frac{d}{dx}(cos2x)+cos2x\frac{d}{dx}(log5x)]$
$⇒\frac{dy}{dx}=3y[log5x(-sin2x).\frac{d}{dx}(2x)+cos2x.\frac{1}{5x}.\frac{d}{dx}(5x)]$
$⇒\frac{dy}{dx}=3y[-2sin2xlog5x+\frac{cos2x}{x}]$
$⇒\frac{dy}{dx}=3y[\frac{3cos\,2x}{x}-6sin\,2x\,log\,5x]$
$∴\frac{dy}{dx}=(5x)^{3cos2x}[\frac{3cos\,2x}{x}-6sin\,2x\,log\,5x]$