# How do you use the fundamental theorem of calculus to find F'(x) given F(x)=int 1/t^2dt from [1,x]?

Mar 22, 2018

$F ' \left(x\right) = \frac{1}{x} ^ 2$

#### Explanation:

The first part of the Fundamental Theorem of Calculus tells us that if

$F \left(x\right) = {\int}_{a}^{x} f \left(t\right) \mathrm{dt}$ where $a$ is just a constant, then $F ' \left(x\right) = f \left(x\right)$.

This makes sense -- $F \left(x\right)$ is an integral, differentiating $F \left(x\right)$ to get $F ' \left(x\right)$ should just give us what we originally integrated, IE, the integrand $F \left(t\right)$ evaluated from a constant to $x$.

The most important thing to note is that the variable of the derivative and variable of the integrand are different. The integrand is written in terms of $t ,$ the integral and derivative in terms of $x .$

Here, we have

$F \left(x\right) = {\int}_{1}^{x} \frac{1}{t} ^ 2 \mathrm{dt}$

And we see $a = 1$ (the value of $a$ is totally irrelevant to our final answer, noting it anyways), $f \left(t\right) = \frac{1}{t} ^ 2.$

Thus,

$F ' \left(x\right) = \frac{1}{x} ^ 2$