Differentiate $x^{x\cos x}+\dfrac{x^{2}+1}{x^{2}-1}$ with respect to $x$.

$x^{x\cos x}\big[\cos x(1+\log x)-x\sin x\log x\big]-\dfrac{4x}{(x^{2}-1)^{2}}$

CBSE Class 12 · 2011 · DifferentiationLong Answer

Differentiate

x^{x c o s x} + \frac{x ^{2} + 1}{x ^{2} - 1}

with respect to

x

Answer: $x^{x c o s x} [cos x (1 + lo g x) - x sin x lo g x] - \frac{4 x}{( x ^{2} - 1 ) ^{2}}$

Let

y = u + v

with

u = x^{x c o s x}, v = \frac{x ^{2} + 1}{x ^{2} - 1}

.
For

u

lo g u = x cos x lo g x

; differentiate the triple product on the right:

\frac{1}{u} \frac{d u}{d x} = cos x lo g x - x sin x lo g x + cos x

\frac{d u}{d x} = x^{x c o s x} [cos x (1 + lo g x) - x sin x lo g x]

For

v

(quotient rule):

\frac{d v}{d x} = \frac{( x ^{2} - 1 ) 2 x - ( x ^{2} + 1 ) 2 x}{( x ^{2} - 1 ) ^{2}} = \frac{- 4 x}{( x ^{2} - 1 ) ^{2}}

Add:

\frac{d y}{d x} = x^{x c o s x} [cos x (1 + lo g x) - x sin x lo g x] - \frac{4 x}{( x ^{2} - 1 ) ^{2}}

CBSE Class 12 Board · 2011 · Subjective