If #f(x)=int_(2x)^sinxcos(t^3)dt# then #f'(x)# is equal to?

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Answer 1 · 2018-04-09T17:11:36

#cos x cos (sin^3x) - 2cos(8x^3)#

One can use the result

#d/dx (int_{a(x)}^{b(x)} G(t)dt) = b^'(x) G(b(x))-a^'(x)G(a(x)) #

to get

#d/dx f(x) = d/dx (int_(2x)^sinx cos(t^3)dt)#
#qquad = cos x cos (sin^3x) - 2cos(8x^3)#

Derivation of the basic result

Let

#F(x) = int_{a(x)}^{b(x)} G(t)dt#

Then

#F(x+Delta x)= int_{a(x+Delta x)}^{b(x+Delta x)} G(t)dt #

so
#F(x+Delta x)-F(x) = int_{a(x+Delta x)}^{b(x+Delta x)} G(t)dt - int_{a(x)}^{b(x)} G(t)dt#

#color(white)(rrr)#

#qquad = int_{a(x+Delta x)}^{a(x)} G(t)dt + int_{b(x)}^{b(x+Delta x)} G(t)dt #

#color(white)(rrr)#

#qquad ~~ - int_{a(x)}^{a(x)+a^'(x)Delta x} G(t)dt+ int_{b(x)}^{b(x)+b^'(x)Delta x} G(t)dt#

#color(white)(rrr)#

#qquad ~~ Delta x[-a^'(x)G(a(x))+b^'(x)G(b(x))]#

and the final result follows from this.