Derivatives Of Trigonometric Functions h(x)=4sinx cosx?

Question

how would I derive this?

h(x)= 4sinx cosx

Answer 1

Take the constant multiple of 4 outside and use the product rule:

Let  f(x) = sin(x)     And let g(x) = cos(x) .   (g ' (x) = derivative of g (x))

Product rule states  y'(x) = f ' (x) . g (x) + f (x) . g ' (x).

in this case    g ' (x) = -sin(x) and  f ' (x) = cos(x)

so h'(x) = 4( cos(x)cos(x) - sin(x)sin(x) )

From there you can use trig identities to get it into a simpler form.

  

Answer 2

I think you have to use the product rule, f'(x) = f'g + fg'

Let f = 4sinx; f' = 4cosx

Let g = cosx; g' = -sinx

f'(x)=f'g + fg'

= 4cosx*cosx + 4sinx*-sinx

=4cos^2x - 4sin^2x

=4(cos^2x - sin^2x)

(I think).