Differentiate the function $\sin x^2 + \sin^2 x + \sin^2 (x^2)$ with respect to $x$.

$2x\cos(x^2) + \sin 2x + 2x\sin 2(x^2)$

The correct answer is Option (2) → $2x\cos(x^2) + \sin 2x + 2x\sin 2(x^2)$ ##

Let $y = \sin x^2 + \sin^2 x + \sin^2 (x^2)$

On differentiating w.r.t. $x$, we get

$\frac{dy}{dx} = \frac{d}{dx} \sin(x^2) + \frac{d}{dx} (\sin x)^2 + \frac{d}{dx} (\sin x^2)^2$

$= \cos(x^2) \cdot \frac{d}{dx}(x^2) + 2 \sin x \cdot \frac{d}{dx} \sin x + 2 \sin x^2 \cdot \frac{d}{dx} \sin x^2 \quad \left[ ∵ \frac{d}{dx} \sin x = \cos x \right]$

$= \cos x^2 \cdot 2x + 2 \cdot \sin x \cdot \cos x + 2 \sin x^2 \cdot \cos x^2 \cdot \frac{d}{dx} x^2$

$= 2x \cos (x)^2 + 2 \cdot \sin x \cdot \cos x + 2 \sin x^2 \cdot \cos x^2 \cdot 2x$

$= 2x \cos (x)^2 + \sin 2x + \sin 2(x)^2 \cdot 2x \quad [∵\sin 2x = 2 \sin x \cos x]$

$= 2x \cos (x^2) + 2x \cdot \sin 2(x^2) + \sin 2x$