Find the derivative of the function given by $f(x) = \sin(x^2)$.

$2x \cos(x^2)$

The correct answer is Option (4) → $2x \cos(x^2)$ ##

Observe that the given function is a composite of two functions. Indeed, if $t = u(x) = x^2$ and $v(t) = \sin t$, then

$f(x) = (v \circ u)(x) = v(u(x)) = v(x^2) = \sin x^2$

Put $t = u(x) = x^2$. Observe that $\frac{dv}{dt} = \cos t$ and $\frac{dt}{dx} = 2x$ exist. Hence, by chain rule

$\frac{df}{dx} = \frac{dv}{dt} \cdot \frac{dt}{dx} = \cos t \cdot 2x$

It is normal practice to express the final result only in terms of $x$. Thus

$\frac{df}{dx} = \cos t \cdot 2x = 2x \cos x^2$