Question

How do you use the product rule to differentiate `sin(x^2)(cos(x^2))`?

Answer 1

`2xcos^2(x^2)-2xsin^2(x^2)`

Explanation:

#d/dx[sin(x^2)cos(x^2)]=(sin(x^2))(d/dx[cos(x^2)])+(d/dx[sin(x^2)])(cos(x^2))= (sin(x^2))(-2xsin(x^2))+(2xcos(x^2))(cos(x^2))=2xcos^2(x^2)-2xsin^2(x^2)#

Don't forget to chain rule `x^2`

Answer 2

`2x*cos(2*x^2)`

Explanation:

Let `f(x) = sin(x^2)` and `g(x) = cos(x^2)`

You have :
`f'(x) = 2x*cos(x^2)`
and
`g'(x) = -2x*sin(x^2)`

then you apply the formula :
`f'(x)*g(x) + f(x)*g'(x)`

which gives you :
`2x*(cos^2(x^2)-sin^2(x^2))`
Hence, with the formula `cos(2a) = cos^2 a - sin^2 a`,
`2x*cos(2*x^2)`

Answer 3

The differentiation of a product is determined by applying this method:
`" "`
`d/dx (f (x)xxg (x))=d/dxf (x)xxg(x)+d/dxg (x)xxf (x)`
`" "`
`(sin (x^2)(cos (x^2)))'`
`" "`
`(sin (x^2))'xxcos (x^2)+(cosx^2)'xxsinx^2`
`" "`
`=2xcosx^2xxcosx^2-2xsinx^2xxsinx^2`
`" "`
`=2x (cosx^2xxcosx^2-sinx^2xxsinx^2)`
`" "`
`=2x (cos^2 (x^2)-sin^2 (x^2))`
`" "`
`=2xcos (2x^2)`