Find $\dfrac{dy}{dx}$, if $y=(\cos x)^{x}+(\sin x)^{1/x}$.

$(\cos x)^{x}\big[\log\cos x-x\tan x\big]+(\sin x)^{1/x}\!\left[\dfrac{\cot x}{x}-\dfrac{\log\sin x}{x^{2}}\right]$

CBSE Class 12 · 2010 · DifferentiationLong Answer

Find

\frac{d y}{d x}

, if

y = (cos x)^{x} + (sin x)^{1/ x}

Answer: $(cos x)^{x} [lo g cos x - x tan x] + (sin x)^{1/ x} [\frac{cot x}{x} - \frac{lo g sin x}{x ^{2}}]$

Let

y = u + v

with

u = (cos x)^{x}, v = (sin x)^{1/ x}

.
For

u

lo g u = x lo g cos x

, so

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

For

v

lo g v = \frac{1}{x} lo g sin x

, so

\frac{d v}{d x} = (sin x)^{1/ x} [- \frac{1}{x ^{2}} lo g sin x + \frac{cot x}{x}]

Add:

\frac{d y}{d x} = (cos x)^{x} [lo g cos x - x tan x] + (sin x)^{1/ x} [\frac{cot x}{x} - \frac{lo g sin x}{x ^{2}}]

CBSE Class 12 Board · 2010 · Subjective