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If \( y=(x)^x+(\cos x)^{2x} \: find \: \large\frac{dy}{dx}\)

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  • $\large\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$
Step 1:
$y=x^x+(\cos x)^{2x}$
Let $y=y_1+y_2$
Where $y_1=x^x$
$y_2=(\cos x)^{2x}$
Consider $y_1=x^x$
Take $\log$ on both sides
$\log y_1=x\log x$
Now differentiating w.r.t $x$ we get,
$\large\frac{1}{y_1}\frac{dy}{dx}$$=x.\large\frac{1}{x}$$+\log x.1$
$\large\frac{dy}{dx}$$=y_1[1+\log x]$
$\quad=x^x[1+\log x]$-----(1)
Step 2:
Consider $y_2=(\cos x)^{2x}$
Take $\log$ on both sides
$\log y_2=2x.\log(\cos x)$
Differentiating w.r.t $x$ we get,
$\large\frac{1}{y_2}\frac{dy}{dx}$$=2x.\large\frac{1}{\cos x}$$(-\sin x)+\log (\cos x)$
$\therefore \large\frac{dy}{dx}$$=y_2[\large\frac{-2x\sin x}{\cos x}$$+2\log(\cos x)]$
$\qquad=(\cos x)^{2x}[2\log x(\cos x)-2x.\tan x]$-----(2)
Step 3:
Combining equ(1) and equ(2) we get,
$\large\frac{dy}{dx}$$=x^x[1+\log x]+(\cos x)^{2x}[2\log(\cos x)-2x\tan x]$
answered Sep 25, 2013 by sreemathi.v
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