#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#