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

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Answer 1 · 2016-09-08T09:26:26

$=0$

this calls for the 2nd Fundamental Theorem of Calculus:

$d/(dz) int_a^z f(t) dt = f(z)$

We are going to do:

$d/dx (color(red)(int_0^y e^t dt) + color(blue)(int_0^x cos(t^2) dt) )$

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

$= e^y * dy/dx + cos(x^2)$

$= d/dx (0) = 0$

int_0^y e^t dtint_0^y e^t dt + int_0^x cos(t^2) dtint_0^x cos(t^2) dt= 0 Then find the value of e^y*dy/dx + cos(x^2)e^y*dy/dx + cos(x^2) ?