यदि y = x^cosx + (cosx)^sinx तो dy/dx ज्ञात करें। ​

Question

यदि y = x^cosx + (cosx)^sinx तो \(\frac{dy}{dx}\) ज्ञात करें।

Answer 1

माना \(y = u +v\)

जहाँ \(u = x^{\cos x}\) और \(v = (\cos x)^{\sin x}\)

अब,

\(u = x^{\cos x}\)

⇒ \(\log u = \cos x\log x\)

⇒ \(\frac 1u \frac{du}{dx} = \frac 1x\cos x - \sin x \log x\)

⇒ \(\frac {du}{dx} = u\left[ \frac{\cos x}{x} - \sin x.\log x\right]\)

⇒ \(\frac {du}{dx} =x^{\cos x}\left[ \frac{\cos x}{x} - \sin x.\log x\right]\)

और \(v = (\cos x)^{\sin x}\)

⇒ \(\log v = \sin x.\log \cos x\)

⇒ \(\frac 1v \frac{dv}{dx} = \sin x \frac{d(\log \cos x)}{dx} + \log \cos x. \frac{d(\sin x)}{dx}\)

\(= \sin x.\frac {(1 -\sin x)}{\cos x} + (\log \cos x) \cos x\)

⇒ \(\frac{dv}{dx} = v [\cos (\log \cos x) - \sin x. \tan x]\)

⇒ \(\frac{dv}{dx} = (\cos x)^{\sin x} [\cos (\log \cos x) - \sin x. \tan x]\)

\(\therefore y = u + v\)

तब,

\(\frac{dy}{dx}= \frac{du}{dx} + \frac{dv}{dx}\)

\(=x^{\cos x}\left[ \frac{\cos x}{x} - \sin x.\log x\right]+ (\cos x)^{\sin x} [\cos (\log \cos x) - \sin x. \tan x]\)