If $y=e^{x^{2}\cos x}+(\cos x)^{x}$, then find $\dfrac{dy}{dx}$.

$e^{x^{2}\cos x}\big(2x\cos x-x^{2}\sin x\big)+(\cos x)^{x}\big[\log\cos x-x\tan x\big]$

CBSE Class 12 · 2020 · DifferentiationLong Answer

y = e^{x^{2} c o s x} + (cos x)^{x}

, then find

\frac{d y}{d x}

Answer: $e^{x^{2} c o s x} (2 x cos x - x^{2} sin x) + (cos x)^{x} [lo g cos x - x tan x]$

Let

y = e^{x^{2} c o s x} + u

with

u = (cos x)^{x}

.
First term by the chain rule, differentiating the exponent

x^{2} cos x

by the product rule:

\frac{d}{d x} e^{x^{2} c o s x} = e^{x^{2} c o s x} (2 x cos x - x^{2} sin x)

For

u = (cos x)^{x}

lo g u = x lo g cos x

, so

\frac{d u}{d x} = (cos x)^{x} [lo g cos x - x tan x]

Add:

\frac{d y}{d x} = e^{x^{2} c o s x} (2 x cos x - x^{2} sin x) + (cos x)^{x} [lo g cos x - x tan x]

CBSE Class 12 Board · 2020 · Subjective