# How do you use the second fundamental theorem of Calculus to find the derivative of given int 3/t^2 dt from [1,x]?

Jul 2, 2016

$\frac{d}{\mathrm{dx}} {\int}_{1}^{x} \frac{3}{t} ^ 2 \mathrm{dt} = \frac{3}{x} ^ 2$

#### Explanation:

you want $\frac{d}{\mathrm{dx}} {\int}_{1}^{x} \frac{3}{t} ^ 2 \mathrm{dt}$

the second fundie theorem states that
$\frac{d}{\mathrm{dx}} {\int}_{a}^{x} f \left(t\right) \mathrm{dt} = f \left(x\right)$

so here simply by pattern matching we get

$\frac{d}{\mathrm{dx}} {\int}_{1}^{x} \frac{3}{t} ^ 2 \mathrm{dt} = \frac{3}{x} ^ 2$