Question

How do you differentiate `y = 3x cos^2 (x)`?

Answer 1

`3cosx`[`cosx-2xsinx`]

Explanation:

`d`[`uv`]= `vdu+udv` this is the product rule, where v and u are both functions of x.

Let u =`3x` and `v=cos^2x`

so we have ......`cos^2x`.[`3]`+`3x`[`-2sinxcosx`]

=`3cos^2x`-`6xsinxcosx`.....=`3cosx[cosx-2xsinx`].

`cos^2x`=`[cosx]^2` and so need to use the chain rule to differentiate this, =`2cosx`, times the derivative of `cosx` which is -`sinx`, so`d/dxcos^2x` `=-2sinxcosx`. hope this helped.