Question

How do you find the derivative of `f(x) = (sin x)(cos x)`?

Answer 1

f'(x) = cos2x

Explanation:

differentiate using the `color(blue)"product rule"`

`color(red)(|bar(ul(color(white)(a/a)color(black)(f(x)=g(x)h(x)rArrf'(x)=g(x)h'(x)+h(x)g'(x))color(white)(a/a)|)))`

`color(orange)"Reminder"`

`color(red)(|bar(ul(color(white)(a/a)color(black)(d/dx(sinx)=cosx" and " d/dx(cosx)=-sinx)color(white)(a/a)|)))`

here `g(x)=sinxrArrg'(x)=cosx`

and `h(x)=cosxrArrh'(x)=-sinx`

Substitute these values into f'(x)

`rArrf'(x)=cosx(-sinx)+cosx(cosx)`

`=cos^2x-sin^2x`

`color(orange)"Reminder" color(red)(|bar(ul(color(white)(a/a)color(black)(cos2x=cos^2x-sin^2x)color(white)(a/a)|)))`

`rArrf'(x)=cos^2x-sin^2x=cos2x`
`color(magenta)"----------------------------------------------------------"`

Alternatively

`color(orange)"Reminder" color(red)(|bar(ul(color(white)(a/a)color(black)(sin2x=2sinxcosx)color(white)(a/a)|)))`

and recognise that `f(x)=sinxcosx=1/2sin2x`

differentiate f(x) using the `color(blue)"chain rule"`

`rArrf'(x)=1/2cos2x.d/dx(2x)=1/2cos2x.2=cos2x`

Answer 2

`(d f(x))/(d x) (sin x)( cos x)=1-2 sin^2 x=cos 2x`

Explanation:

`(d f(x))/(d x) (sin x)( cos x)=?`

`d/(d x) sin x=cos x`

`d/(d x)cos x=-sin x`

`y=a*b" ; "y^'=a^'*b+b^'*a`

`(d f(x))/(d x) (sin x)( cos x)=cos x*cos x-sin x*sin x`

`(d f(x))/(d x) (sin x)( cos x)=cos^2 x-sin^2 x`

`"remember that :"cos^2 x=1-sin^2 x`

`(d f(x))/(d x) (sin x)( cos x)=1-sin^2 x-sin^2 x`

`(d f(x))/(d x) (sin x)( cos x)=1-2 sin^2 x=cos 2x`