Browse Questions

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

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)$