Q 11   Differentiate the functions w.r.t. x.   ( x \cos x )^ x + ( x \sin x )^{1/ x}

Answers (1)

Given function is
f(x)=( x \cos x )^ x + ( x \sin x )^{1/ x}
Let's take t = (x\cos x)^x
Now, take log on both sides 
\log t =x\log (x\cos x) = x(\log x+\log \cos x)
Now, differentiate w.r.t. x
we get,
\frac{1}{t}\frac{dt}{dx} =(\log x+\log \cos x)+x(\frac{1}{x}+\frac{1}{\cos x}.(-\sin x))\\ \frac{dt}{dx} = t(\log x + \log \cos x+1-x\tan x) \ \ \ \ \ \ \ \ \ (\because \frac{\sin x}{\cos x}= \tan x)\\ \frac{dt}{dx}= (x\cos x)^x(\log x + \log \cos x+1-x\tan x)\\ \frac{dt}{dx}=(x\cos x)^x(+1-x\tan x+\log (x\cos x))
Similarly, take k = (x\sin x)^{\frac{1}{x}}
Now, take log on both the sides
\log k = \frac{1}{x}(\log x+\log \sin x)
Now, differentiate w.r.t. x
we get,
\frac{1}{k}\frac{dk}{dx} =(\frac{-1}{x^2})(\log x+\log \sin x)+\frac{1}{x}(\frac{1}{x}+\frac{1}{\sin x}.(\cos x))\\ \frac{dk}{dx} = \frac{k}{x^2}(-\log x - \log \sin x+\frac{1}{x^2}+\frac{\cot x}{x}) \ \ \ \ \ \ \ \ \ (\because \frac{\cos x}{\sin x}= \cot x)\\ \frac{dk}{dx}=\frac{(x\sin x)^{\frac{1}{x}}}{x^2}(-\log x - \log \sin x+\frac{1}{x^2}+\frac{\cot x}{x})\\ \frac{dk}{dx}=(x\sin x)^{\frac{1}{x}}\frac{(x\cot x+1-(\log x\sin x))}{x^2}
Now,
\frac{dy}{dx}= \frac{dt}{dx}+\frac{dk}{dx}
\frac{dy}{dx}= (x\cos x)^x(+1-x\tan x+\log (x\cos x))+(x\sin x)^{\frac{1}{x}}\frac{(x\cot x+1-(\log x\sin x))}{x^2}
Therefore, the answer is (x\cos x)^x(1-x\tan x+\log (x\cos x))+(x\sin x)^{\frac{1}{x}}\frac{(x\cot x+1-(\log x\sin x))}{x^2}
 

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