How do you differentiate #f(x)=e^x*sin(5x^2)# using the product rule?

2 Answers
Apr 28, 2018

Answer:

#e^x(sin(5x^2)+10xcos(5x^2))#

Explanation:

Let #u=e^x# and #v=sin(5x^2)#
Therefore #du/dx=e^x#
and to find #dv/dx# you need to use chain rule...
so... let #w=5x^2# then #v=sin(w)#
#dv/dw = cos(w)#
and #dw/dx=10x#
so the differential of #v=sin(5x^2)# is #10xcos(5x^2)#

from there you use the rest of product rule so... #dy/dx=(v*du/dx)+(u*dv/dx)#
so...
#(sin(5x^2)*e^x)+(e^x*10xcos(5x^2))#
#e^x(sin(5x^2)+10xcos(5x^2))#

Apr 28, 2018

Answer:

#f'(x)=e^x(10xcos(5x^2)+sin(5x^2))#

Explanation:

#"Given "f(x)=g(x)h(x)" then"#

#f'(x)=g(x)h'(x)+h(x)g'(x)larrcolor(blue)"product rule"#

#g(x)=e^xrArrg'(x)=e^x#

#h(x)=sin(5x^2)larrcolor(blue)"use chain rule"#

#rArrh'(x)=cos(5x^2)xxd/dx(5x^2)#

#color(white)(rArrh'(x))=10xcos(5x^2)#

#rArrf'(x)=10xcos(5x^2)e^x+sin(5x^2)e^x#

#color(white)(rArrf'(x))=e^x(10xcos(5x^2)+sin(5x^2))#