find dy/dx of (xcosx)^x+(xsinx)^1/x - Maths - Continuity and Differentiability

Question

find dy/dx of (xcosx)^x+(xsinx)^1/x

Santosh Kumar · Accepted Answer

let y = (x cos x)x + (x sin x)1/x let u = (x cos x)x and v = (x sin x)1/x Then, log u = x log (x cos x) â‡’log u = x [log x + log (cos x)] $\frac{1}{u} \cdot \frac{d u}{d x} = x \frac{d}{d x} [\log x + \log (\cos x)] + [\log x + \log (\cos x)] \frac{d x}{d x} \frac{1}{u} \cdot \frac{d u}{d x} = \frac{x}{x} + \frac{x}{\cos x} (- \sin x) + \log x + \log (\cos x) \Rightarrow \frac{1}{u} \cdot \frac{d u}{d x} = 1 - x \tan x + \log x + \log (\cos x) \Rightarrow \frac{d u}{d x} = {(x \cos x)}^{x} [1 - x \tan x + \log x + \log (\cos x)] (i)$ Also, v = (x sin x)1/x $\Rightarrow \log v = \frac{1}{x} \log (x \sin x) \Rightarrow \log v = \frac{1}{x} [\log x + \log (\sin x)] \frac{1}{v} \cdot \frac{d v}{d x} = \frac{1}{x} \cdot \frac{d}{d x} [\log x + \log (\sin x)] + [\log x + \log (\sin x)] \frac{d}{d x} (\frac{1}{x})$ $\Rightarrow \frac{1}{v} \cdot \frac{d v}{d x} = \frac{1}{x} \cdot \frac{1}{x} + \frac{1}{x} \cdot \frac{\cos x}{\sin x} - [\log x + \log (\sin x)] \frac{1}{x^{2}} \Rightarrow \frac{d v}{d x} = (x \sin x)^{1 / x} [\frac{1}{x^{2}} + \frac{1}{x} \cdot \cot x - \frac{1}{x^{2}} \log x - \frac{1}{x^{2}} \log (\sin x)] (i i)$ Now, $\frac{d y}{d x} = \frac{d u}{d x} + \frac{d v}{d x} [from equation (i) and (i i)] \Rightarrow \frac{d y}{d x} = (x \cos x)^{x} [1 - x \tan x + \log x + \log (\cos x)] + (x \sin x)^{1 / x} [\frac{1}{x^{2}} + \frac{1}{x} \cdot \cot x - \frac{1}{x^{2}} \log x - \frac{1}{x^{2}} \log (\sin x)]$ $\Rightarrow \frac{d y}{d x} = (x \cos x)^{x} [1 - x \tan x + \log (x \cos x)] + (x \sin x)^{1 / x} [\frac{1}{x^{2}} + \frac{1}{x} \cdot \cot x - \frac{1}{x^{2}} \log (x \sin x)]$