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Differentiate the functions given in w.r.t. $x : $ $ x^{\large x} - 2^{\large sin \: x} $

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Toolbox:
  • $\large\frac{dy}{dx}=\frac{du}{dx}-\frac{dv}{dx}$
  • $\log m^n=n\log m$
Step 1:
Let $y=x^{\large x}-2^{\large\sin x}$
$\qquad=u-v$
Now $u=x^{\large x}$
Taking $\log$ on both sides
$\log u=\log x^{\large x}$
$\log u=x\log x$
$\log m^n=n\log m$
Step 2:
Differentiating on both sides
$\large\frac{1}{u}\frac{du}{dx}$$=1.\log x+x.\large\frac{1}{x}$
$\qquad\;=\log x+1$
$\large\frac{du}{dx}$$=u(1+\log x)$
$\qquad=x^{\large x}(1+\log x)\quad [Replacing\; u=x^{\large x}$]
Step 3:
Taking $v=2^{\large\sin x}$
Taking $\log$ on both sides
$\log v=\log 2^{\large\sin x}$
$\qquad=\sin x\log 2$
$\large\frac{1}{v}\frac{dv}{dx}$$=\cos x\log 2$
$\large\frac{dv}{dx}$$=v\cos x\log 2$
$\quad=2^{\large\sin x}\cos x \log 2$
Step 4:
$\large\frac{dy}{dx}=\frac{du}{dx}-\frac{dv}{dx}$
$\quad=x^{\large x}(1+\log x)-2^{\large\sin x}\cos x\log 2$
answered May 8, 2013 by sreemathi.v
 

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