Question

Differentiate (cos⁡3x)^3x with respect to x.

(a) (cos⁡3x)^x (3 log⁡(cos⁡3x) – 9x tan⁡3x)

(b) (cos⁡3x)^3x (3 log⁡(cos⁡3x) + 9x tan⁡3x)

(c) (cos⁡3x)^3x (3 log⁡(cos⁡3x) – 9x tan⁡3x)

(d) (cos⁡3x)^3x (log⁡(cos⁡3x) + 9 tan⁡3x)

I had been asked this question during an interview.

My question comes from Logarithmic Differentiation in division Continuity and Differentiability of Mathematics – Class 12

Answer 1

Correct option is (c) (cos⁡3x)^3x (3 log⁡(cos⁡3x) – 9x tan⁡3x)

The best explanation: Consider y=(cos⁡3x)^3x

Applying log on both sides, we get

log⁡y=log⁡(cos⁡3x)^3x

log⁡y=3x log⁡(cos⁡3x)

Differentiating both sides with respect to x, we get

$\frac{1}{y} \frac{dy}{dx}=\frac{d}{dx} (3x \,log⁡(cos⁡3x))$

By using u.v=u’ v+uv’, we get

$\frac{1}{y} \frac{dy}{dx}$=$\frac{d}{dx} \,(3x) \,log⁡(cos⁡3x)+\frac{d}{dx} \,(log⁡(cos⁡3x)).3x$

$\frac{dy}{dx}$=y(3 log⁡(cos⁡3x) + $\frac{1}{cos⁡3x} \,. \frac{d}{dx} \,(cos⁡3x).3x)$

$\frac{dy}{dx}$=y(3 log⁡(cos⁡3x) + $\frac{1}{cos⁡3x} \,. \,(-sin⁡3x).\frac{d}{dx}(3x).3x)$

$\frac{dy}{dx}$=y(3 log⁡(cos⁡3x) + $\frac{1}{cos⁡3x} \,. \,(-sin⁡3x).3.3x)$

$\frac{dy}{dx}$=y(3 log⁡(cos⁡3x) – 9x tan⁡3x)

$\frac{dy}{dx}$=(cos⁡3x)^3x (3 log⁡(cos⁡3x) – 9x tan⁡3x)