Question

How do you differentiate `f(x)=sinx-4cos(5x)`?

Answer 1

`f'(x)=\cos x+20\sin(5x)`

Explanation:

The given function:

`f(x)=\sin x-4\cos(5x)`

Differentiating above function w.r.t. `x` using chain rule as follows

`d/dxf(x)=d/dx(\sin x-4\cos(5x))`

`f'(x)=d/dx(\sin x)-4\d/dx (cos(5x))`

`f=cos x-4(-sin(5x))d/dx(5x)`

`=\cos x+4\sin(5x)(5)`

`=\cos x+20\sin(5x)`

Answer 2

`cos(x)+20sin(5x)`

Explanation:

`f(x)=sin(x)-4cos(5x)`

So

`f'(x)=cos(x)-4(-sin(5x) * 5)`

`f'(x)=cos(x)+20sin(5x)`

Hope it helps!

Answer 3

`cosx+20sin(5x)`

Explanation:

We essentially have the following:

`color(steelblue)(d/dx sinx)-color(purple)(d/dx 4cos(5x))`

What I have in blue evaluates to `cosx`. We now have

`cosx-4d/dxcolor(purple)(cos(5x))`

What I have in purple is a composite function with `f(x)=cosx` and `g(x)=5x`. We can find the derivative with the Chain Rule

`f'(g(x))*g'(x)`

We know:

`f(x)=cosx=>f'(x)=-sinx`
`g(x)=5x=>g'(x)=5`

We can plug this into the Chain Rule to get

`color(purple)(-5sin(5x))`

We now have the following:

`cosx-4color(purple)((-5sin(5x))`, which simplifies to

`cosx+20sin(5x)`

Hope this helps!