Question

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

Answer 1

`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)`