Question

Differentiate the following with respect to x. 

(i) x^{sin x} 

(ii) (sin x)^x

(iii) (sin x)^{tan x} 

(iv) \(\sqrt{\frac{(x-1)(x-2)}{(x-3)(x^2+x+1)}}\)

Answer 1

(i) Let y = x^{sin x} 

Taking logarithm on both sides we get, 

log y = log(x^{sin x}) 

log y = sin x log x 

Differentiating with respect to x,

(ii) Let y = (sin x)^x 

Taking logarithm on both sides we get, 

log y = x log(sin x) 

Differentiating with respect to x, 

\(\frac{1}{y}.\frac{dy}{dx}\) = x \(\frac{d}{dx}\)log(sin x) + log(sin x) \(\frac{d}{dx}\)(x) 

\(\frac{1}{y}.\frac{dy}{dx}\) = x \(\frac{1}{sin\,x}\)(cos x) + log(sin x) (1) 

\(\frac{1}{y}.\frac{dy}{dx}\) = x cot x + log(sin x) 

\(\frac{dy}{dx}\) = y[x cot x + log(sin x)] 

\(\frac{dy}{dx}\) = (sin x)^x [x cot x + log(sin x)]

(iii) Let y = (sin x)^{tan x} 

Taking logarithm on both sides we get, 

log y = tan x log(sin x)

Differentiating with respect to x,

(iv) Let y =\(\sqrt{\frac{(x-1)(x-2)}{(x-3)(x^2+x+1)}}\)

y = \(({\frac{(x-1)(x-2)}{(x-3)(x^2+x+1)}})^\frac{1}{2}\)

Taking logarithm on both sides we get, 

log y = \(\frac{1}{2}\){[log(x – 1) + log(x – 2)] – [(log(x – 3) + log(x² + x + 1)]} 

log y = \(\frac{1}{2}\)[log(x – 1) + log(x – 2) – log(x – 3) – log(x² + x + 1)]

Differentiating with respect to x,