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Differentiate w.r.t. \(x\) the function in \( sin^3x + cos^6x \)

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  • $\large\frac{dy}{dx}=\frac{du}{dx}$$+ \large\frac{dv}{dx}$
Step 1:
$y=\sin^2x+\cos^6x$
It is of the form $y=u+v$
$\large\frac{dy}{dx}=\frac{du}{dx}$$+ \large\frac{dv}{dx}$
$u=\sin^2x$
Put $\sin x=t$
$u=t^3$
$t=\sin x$
$\large\frac{du}{dt}$$=3t^2$
$\large\frac{dt}{dx}$$=\cos x$
Step 2:
$\large\frac{du}{dx}=\frac{du}{dt}$$\times \large\frac{dt}{dx}$
$\quad\;=3t^2\times \cos x$
Put $t=\sin x$
$\quad\;=3\sin^2 x \cos x$
$\large\frac{du}{dx}$$=3\sin^2 x \cos x$
Step 3:
Now $v=s^{\large 6}$
$s=\cos x$
Differentiating with respect to $s$
$\large\frac{dv}{ds}$$=6s^5$
$s=\cos x$
Differentiating with respect to $x$
$\large\frac{ds}{dx}$$=-\sin x$
Step 4:
$\large\frac{dv}{dx}=\frac{dv}{ds}.\frac{ds}{dx}$
$\quad\;=6s^5(-\sin x)$
$\quad\;=-6.(\cos x)^5\sin x$
$\quad\;=6.\cos^5 x\sin x$
$\large\frac{dy}{dx}=$$3\sin^2x\cos x-6\cos^5x\sin x$
$\quad\;=3\sin x\cos x(\sin x-2\cos^4x)$
answered May 14, 2013 by sreemathi.v
 

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