Question

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

Answer 1

`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)`