if f(x) = ∫_{-1}^{x^2} cos(t) dt, then f(x) = sin(x^2) cos(x^2) 2x cos(x^2) 2x sin(x^2)

Answer:

C. $2x\cos(x^{2})$

Explanation:

Step1: Apply the fundamental theorem of calculus and chain - rule

Let $u = x^{2}$, and $F(t)$ be an antiderivative of $\cos(t)$, i.e., $F^\prime(t)=\cos(t)$. Then $f(x)=\int_{-1}^{x^{2}}\cos(t)dt=F(x^{2})-F(-1)$.

Step2: Differentiate $f(x)$ with respect to $x$

By the chain - rule, $f^\prime(x)=\frac{d}{dx}[F(x^{2})-F(-1)]$. Since $F(-1)$ is a constant, its derivative is 0. And $\frac{d}{dx}F(x^{2})=F^\prime(x^{2})\cdot\frac{d}{dx}(x^{2})$.

Step3: Calculate the derivatives

We know that $F^\prime(t)=\cos(t)$, so $F^\prime(x^{2})=\cos(x^{2})$, and $\frac{d}{dx}(x^{2}) = 2x$. Then $f^\prime(x)=2x\cos(x^{2})$.