Find $F ' (x)$. $F(x) = \int_5^{x^2} \frac{1}{t^4} dt$ First, off. I apologize if my Mathjax is bad. Still learning and it took me a while to write this mathjax.
Question: I can't solve this problem. I understand how $\frac{1}{t^4}$ = $t^-4$ , but I am stuck on what would be next. Can I get guidance? Would it be:
$F(x) = \int_5^{x^2} \frac{1}{t^4} dt$ 
$F(x)= (x^2)^4) - (5)^4)$     
^I could not put a -4 exponent with my mathjax skills yet.
 A: Second FTC states that for an integrable function $f:[a,b]\to \mathbb{R}$ and $x\in]a,b[$ we have
$$G(x)=\int_a^xf(t)dt\quad\implies\quad G'(x)=f(x)$$
From the Chain Rule it follows
$$F(x)=\int_5^{x^2}\frac1{t^4}dt\quad \implies \quad F'(x)=\frac1{(x^2)^4}\cdot(x^2)'=\frac{2x}{x^8}=\frac2{x^7}$$
A: Guide:
Use the Fundamnetal Theorem of Calculus:


*

*If $G(x) = \int_a^x g(t) \, dt$, then $G'(x) = g(x)$.


Also, use chain rule since the upper limit is $x^2$.
A: $\dfrac{d}{dx}\displaystyle\int_{5}^{x^{2}}\dfrac{1}{t^{4}}dt=\dfrac{d}{dx^{2}}\left(\displaystyle\int_{5}^{x^{2}}\dfrac{1}{t^{4}}dt\right)\dfrac{dx^{2}}{dx}=\dfrac{1}{(x^{2})^{4}}\cdot2x=\dfrac{2}{x^{7}}$.
A: For
$$F(x) = \int_5^{x^2} \frac{1}{t^4} dt$$
one can evaluate the integral and then differentiate as follows:
\begin{align}
F(x) &= \int_{5}^{x^{2}} \frac{dt}{t^{4}} \\
&= \left[ - \frac{1}{3 \, t^{3}} \right]_{5}^{x^{2}} \\
&= - \frac{1}{3} \, \left( \frac{1}{x^{6}} - \frac{1}{125} \right). 
\end{align}
Now, 
$$F'(x) = \frac{2}{x^{7}}.$$
