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Differentiate \(( x \cos \: x )^x + ( x\sin\: x )^{\Large\frac{1}{x}}\) w.r.t. \( x \).

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  • $\large\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$
Step 1:
Let $y= (x \cos x )^{\large x} + (x\sin x)^{\Large\frac{1}{x}} $
Its of the form $u+v$
Now $u=(x\cos x)^{\large x}$
Taking $\log$ on both sides
$\log u=\log (x\cos x)^{\large x}$
We know that $\log m^{\large n}=n\log m$
$\log u=x\log(x\cos x)$
$\qquad\;=x(\log x+\log\cos x)$
Step 2:
Differentiating on both sides
$\large\frac{1}{u}\frac{du}{dx}=$$1.[\log x+\log\cos x]+x[\large\frac{1}{x}-\large\frac{\sin x}{\cos x}]$
$\qquad\;=\log x(\cos x)+x\begin{bmatrix}\large\frac{\cos x-x\sin x}{x\cos x}\end{bmatrix}$
$\qquad\;=\log x(\cos x)+\begin{bmatrix}\large\frac{\cos x-x\sin x}{\cos x}\end{bmatrix}$
$\large\frac{du}{dx}$$=u\begin{bmatrix}\log x(\cos x)+\large\frac{\cos x-x\sin x}{\cos x}\end{bmatrix}$
$\large\frac{du}{dx}$$=(x\cos x)^{\large x}\begin{bmatrix}\log (x\cos x)+\large\frac{\cos x-x\sin x}{\cos x}\end{bmatrix}$
Step 3:
$v=(x\sin x)^{\Large\frac{1}{x}}$
Taking $\log$ on both sides
$\log v=\log(x\sin x)^{\Large\frac{1}{x}}$
We know that $\log m^{\large n}=n\log m$
$\log v=\large\frac{1}{x}$$\log(x\sin x)$
$\qquad\;=\large\frac{1}{x}$$(\log x+\log\sin x)$
Differentiating both sides
$\large\frac{1}{v}\frac{dv}{dx}=\big(\large\frac{-1}{x^2}\big)$$\log (x\sin x)+\large\frac{\sin x+x\cos x}{x^2\sin x}$
$\large\frac{dv}{dx}=\big(\large\frac{v}{x^2}\big)$$[\large\frac{\sin x+x\cos x}{\sin x}$$-\log(x\sin x)]$
$\large\frac{dv}{dx}=\big(\large\frac{(x\sin x)^{\Large\frac{1}{x}}}{x^2}\big)$$[\large\frac{\sin x+x\cos x}{\sin x}$$-\log(x\sin x)]$
Step 4:
$\large\frac{dy}{dx}$$=(x\cos x)^{\large x}\begin{bmatrix}\log (x\cos x)+\large\frac{\cos x-x\sin x}{\cos x}\end{bmatrix}$+$\large\frac{(x\sin x)^{\Large\frac{1}{x}}}{x^2}$$[\large\frac{\sin x+x\cos x}{\sin x}$$-\log(x\sin x)]$
answered Sep 20, 2013 by sreemathi.v

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