Question

What is the derivative of `sinxcosx`?

Answer 1

`d/dx(sinxcosx) = cos2x`

Explanation:

The product rule can be used to differentiate any function of the form `f(x) = g(x)h(x)`. It states that `color(red)(f'(x) = g'(x)h(x) + g(x)h'(x)`.

The derivative of `sinx` is `cosx` and the derivative of `cosx` is `-sinx`.

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

`f'(x) = cos^2x - sin^2x`

Use the identity `cos2x = cos^2x - sin^2x`:

`f'(x) = cos2x`

Hopefully this helps!

Answer 2

`cos2x`

Explanation:

Since this is the product of 2 functions, differentiate using the `color(blue)"product rule"`

`"Given " f(x)=g(x)h(x)" then"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(f'(x)=g(x)h'(x)+h(x)g'(x))color(white)(2/2)|)))to(A)`

The following `color(blue)"derivatives"` should be known.

`color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(d/dx(sinx)=cosx" and " d/dx(cosx)=-sinx)color(white)(2/2)|)))`

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

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

substitute these values into (A)

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

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

This may be simplified using the `color(blue)"trigonometric identity"`

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

`rArrd/dx(sinxcosx)=cos2x`

Answer 3

`cos2x.`

Explanation:

Let `y=sinxcosx=1/2(2sinxcosx)=1/2sin2x=1/2sinphi,`

where, `phi=2x.`

Here, we first need to know that, `d/dt{kf(t)}=kd/dt{f(t)},`

where, `k` is a costant.

`:. dy/dx=d/dx(1/2sinphi)=1/2d/dx(sinphi)`

We now use the Chain Rule, which states that :

If `y` is a function (fun.) of `phi, and, phi" is a fun. of "x` [ so

that `y` becomes a fun. of `x`], then, `dy/dx=dy/(dphi)(dphi)/dx`

`:. dy/dx=1/2d/dx(sinphi)`

`=1/2d/(dphi)(sinphi)((dphi)/dx)`

`=1/2cosphid/dx(2x)............[because, phi=2x]`

`=(1/2cosphi)(2d/dx(x))`

`=cosphi=cos2x`

Enjoy Maths!