15. $y = int_{x^{3}}^{5} \frac{cos t}{t^{2}+2} dt$

Explanation:

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

Let $F(t)$ be an antiderivative of $\frac{\cos t}{t^{2}+2}$, i.e., $F^\prime(t)=\frac{\cos t}{t^{2}+2}$. Then $y = F(5)-F(x^{3})$.

Step2: Differentiate $y$ with respect to $x$

By the chain - rule, $\frac{dy}{dx}=0 - F^\prime(x^{3})\cdot3x^{2}$. Since $F^\prime(t)=\frac{\cos t}{t^{2}+2}$, substituting $t = x^{3}$ gives $\frac{dy}{dx}=-\frac{\cos(x^{3})}{(x^{3})^{2}+2}\cdot3x^{2}$.

Step3: Simplify the expression

$\frac{dy}{dx}=-\frac{3x^{2}\cos(x^{3})}{x^{6}+2}$.

Answer:

$-\frac{3x^{2}\cos(x^{3})}{x^{6}+2}$