9. $y = \\int_{0}^{x^{2}}e^{t^{2}}dt$

Explanation:

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

Let $u = x^{2}$, and $F(t)$ be an antiderivative of $e^{t^{2}}$, i.e., $F^\prime(t)=e^{t^{2}}$. Then $y=\int_{0}^{x^{2}}e^{t^{2}}dt=F(x^{2})-F(0)$.

Step2: Differentiate with respect to $x$

By the chain - rule, $\frac{dy}{dx}=F^\prime(x^{2})\cdot\frac{d(x^{2})}{dx}$. Since $F^\prime(t) = e^{t^{2}}$, when $t = x^{2}$, $F^\prime(x^{2})=e^{(x^{2})^{2}}$, and $\frac{d(x^{2})}{dx}=2x$.

Step3: Calculate the derivative

$\frac{dy}{dx}=e^{x^{4}}\cdot2x = 2xe^{x^{4}}$.

Answer:

$2xe^{x^{4}}$