question\ngiven $f(x)$ below, find $f(x)$.\n$f(x)=int_{sqrt{x}}^{x^{3}}t^{4}dt$\nprovide your answer…

Explanation:

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

If $F(t)$ is an antiderivative of $t^4$, i.e., $F^\prime(t)=t^4$, then $\int_{a(x)}^{b(x)}t^4dt=F(b(x)) - F(a(x))$. By the chain - rule, the derivative of $y = F(b(x))-F(a(x))$ with respect to $x$ is $y^\prime=F^\prime(b(x))\cdot b^\prime(x)-F^\prime(a(x))\cdot a^\prime(x)$.

Step2: Identify $a(x)$ and $b(x)$ and their derivatives

Here, $a(x)=\sqrt{x}=x^{\frac{1}{2}}$ and $b(x)=x^{3}$. Then $a^\prime(x)=\frac{1}{2}x^{-\frac{1}{2}}$ and $b^\prime(x) = 3x^{2}$. Also, since $F^\prime(t)=t^4$, we have $F^\prime(b(x))=(x^{3})^{4}$ and $F^\prime(a(x))=(x^{\frac{1}{2}})^{4}$.

Step3: Apply the formula

$f^\prime(x)=(x^{3})^{4}\cdot3x^{2}-(x^{\frac{1}{2}})^{4}\cdot\frac{1}{2}x^{-\frac{1}{2}}$. Simplify the expressions: $(x^{3})^{4}=x^{12}$, so $(x^{3})^{4}\cdot3x^{2}=3x^{12 + 2}=3x^{14}$. And $(x^{\frac{1}{2}})^{4}=x^{2}$, so $(x^{\frac{1}{2}})^{4}\cdot\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2}x^{2-\frac{1}{2}}=\frac{1}{2}x^{\frac{3}{2}}$.

Answer:

$3x^{14}-\frac{1}{2}x^{\frac{3}{2}}$