CBSE Class 12 · 2009 · DifferentiationLong Answer

Differentiate the function

x^{c o s x} + (sin x)^{t a n x}

with respect to

x

Answer: $x^{c o s x} [\frac{cos x}{x} - sin x lo g x] + (sin x)^{t a n x} [1 + sec^{2} x lo g sin x]$

Solution

Let

y = u + v

with

u = x^{c o s x}, v = (sin x)^{t a n x}

.
For

u

lo g u = cos x lo g x

, so

\frac{d u}{d x} = x^{c o s x} [\frac{cos x}{x} - sin x lo g x]

For

v

lo g v = tan x lo g sin x

, so

\frac{d v}{d x} = (sin x)^{t a n x} [1 + sec^{2} x lo g sin x]

Add:

\frac{d y}{d x} = x^{c o s x} [\frac{cos x}{x} - sin x lo g x] + (sin x)^{t a n x} [1 + sec^{2} x lo g sin x]

CBSE Class 12 Board · 2009 · Subjective