How do you differentiate #g(x) =sin(3x) (2x+1)^(5/2)# using the product rule?

1 Answer
Jan 19, 2018

Answer:

#g'(x)=5sin(3x)(2x+1)^(3/2)+3cos(3x)(2x+1)^(5/2)#

Explanation:

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

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

#"differentiate "sin(3x)" and "(2x+1)^(5/2)#

#"using the "color(blue)"chain rule"#

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

#dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"#

#d/dx(sin3x)=cos(3x)xxd/dx(3x)=3cos(3x)#

#d/dx((2x+1)^(5/2))=5/2(2x+1)^(3/2).2=5(2x+1)^(3/2)#

#f(x)=sin(3x)tof'(x)=3cos(3x)#

#h(x)=(2x+1)^(5/2)toh'(x)=5(2x+1)^(3/2)#

#rArrg'(x)=5sin(3x)(2x+1)^(3/2)+3cos(3x)(2x+1)^(5/2)#