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Differentiate the following w.r.t $x$ : $\sin (x^2)+\sin^2x+\sin^2(x^2)$.

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Toolbox:
  • Chain Rule : Suppose $f$ is a real value function which is a composite of two functions $u$ and $v$ $(i.e$) $f=uov$ then $\large\frac{df}{dx}=\frac{dv}{dt}$$\times \large\frac{dt}{dx}$
Step 1:
$y=\sin (x^2)+\sin^2x+\sin^2(x^2)$.
Let us consider $y=\sin(x^2)$
On differentiating $\sin(x^2)$ w.r.t $x$ we get $\cos(x^2)$ and on differentiating $x^2$ w.r.t $x$ we get $2x$.
On combining both we get,
$\large\frac{dy}{dx}$$=\cos(x^2).2x$------(1)
Step 2:
Next consider $y=\sin^2x$
On differentiating $\sin^2x$ w.r.t $x$ we get $2\sin x$ and then differentiating $\sin x$ w.r.t $x$ we get $\cos x$.
On combining both we get,
$\large\frac{dy}{dx}$$=2\sin x\cos x=\sin 2x$------(2)
Step 3:
Let us consider $y=\sin^2(x^2)$
On differentiating $\sin^2(x^2)$ w.r.t $x$ we get $2\sin(x^2)$ , on differentiating $\sin x^2$ w.r.t $x$ we get $\cos x^2$ and on differentiating $x^2$ w.r.t $x$,we get $2x$.
Hence on combining we get,
$\large\frac{dy}{dx}$$=2\sin (x^2)\cos x^2. 2x$------(3)
Step 4:
On combining equ(1),(2) and (3) we get,
$\large\frac{dy}{dx}$$=2x\cos x^2+\sin 2x+4x\sin(x)^2.\cos(x)^2$
But $2\sin(x)^2\cos(x)^2=\sin(2x^2)$
Therefore $\large\frac{dy}{dx}$$=2x\cos x^2+\sin 2x+2x\sin(2x^2)$
answered Jun 27, 2013 by sreemathi.v
 
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