f(x)=∫₀ˣ²sin(t)dt\nf(x)=□

Explanation:

Step1: Apply the chain - rule and FTC

Let $u = x^{2}$, then $F(x)=\int_{0}^{u}\sin(t)dt$. By the fundamental theorem of calculus (FTC) and the chain - rule, if $F(x)=\int_{a}^{g(x)}f(t)dt$, then $F^{\prime}(x)=f(g(x))\cdot g^{\prime}(x)$.

Step2: Identify $f$, $g$ and their derivatives

Here, $f(t)=\sin(t)$, $g(x)=x^{2}$, and $g^{\prime}(x) = 2x$.

Step3: Calculate $F^{\prime}(x)$

Substitute $f(g(x))$ and $g^{\prime}(x)$ into the formula. $f(g(x))=\sin(x^{2})$, so $F^{\prime}(x)=\sin(x^{2})\cdot2x = 2x\sin(x^{2})$.

Answer:

$2x\sin(x^{2})$