# int_0^y e^t dt + int_0^x cos(t^2) dt= 0 Then find the value of e^y*dy/dx + cos(x^2) ?

Sep 8, 2016

$= 0$

#### Explanation:

this calls for the 2nd Fundamental Theorem of Calculus:

$\frac{d}{\mathrm{dz}} {\int}_{a}^{z} f \left(t\right) \mathrm{dt} = f \left(z\right)$

We are going to do:

$\frac{d}{\mathrm{dx}} \left(\textcolor{red}{{\int}_{0}^{y} {e}^{t} \mathrm{dt}} + \textcolor{b l u e}{{\int}_{0}^{x} \cos \left({t}^{2}\right) \mathrm{dt}}\right)$

The blue term is straightforward. The red required the use of the Chain Rule too.

$= {e}^{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + \cos \left({x}^{2}\right)$

$= \frac{d}{\mathrm{dx}} \left(0\right) = 0$