How do you differentiate #g(x) =(2x+1)^(5/2)cos(5x) # using the product rule?

1 Answer
Jun 17, 2017

Answer:

#g'(x)=5(2x+1)^(3/2)(cos5x-(2x+1)sin5x)#

Explanation:

#"differentiate using the "color(blue)"product and chain rules"#

#"given " g(x)=f(x).h(x)" then"#

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

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

#f'(x)=g'(h(x)xxh'(x)larr" chain rule"#

#f(x)=(2x+1)^(5/2)rArrf'(x)=5/2(2x+1)^(3/2)xxd/dx(2x+1)#

#color(white)(xxxxxxxxxxxxxxxxxx)=5(2x+1)^(3/2)#

#h(x)=cos5xrArrh'(x)=-sin5x xxd/dx(5x)#

#color(white)(xxxxxxxxxxxxxxxx)=-5sin5x#

#rArrg'(x)=(2x+1)^(5/2)(-5sin5x)+5cos5x(2x+1)^(3/2)#

#color(white)(rArrf'(x))=5(2x+1)^(3/2)(cos5x-(2x+1)sin5x)#