`f(x) = 4cos(5x-2)`

To take the derivative of this, use the chain rule for cosine.

`d/dx(cos u) = -sin u *(du)/dx`

Applying this formula, the derivative of f(x) will be:

`f'(x) = d/dx(4cos (5x - 2))`

`= 4*(-sin(5x - 2)) * d/dx(5x -2)`

`=4*(-sin (5x-2) * 5`

`= -20sin(5x-2)`

**Therefore, `f'(x)=-20sin(5x-2)` .**

f(x)=4 cos ( 5x- 2)

Chain rule dictates we start with the outside function first:

4 is just a multiplier and doesn't matter. The derivative of cosx is -sinx. so:

-4sin(5x - 2)

Then we take the inside function's derivative

so (5x - 2) becomes just 5. (Slope of 5x - 2 is 5)

Then we multiply the two derivatives we just found to get:

-20sin(5x -2)

f(x) = 4cos(5x-2)

To find the drivative of f(x) = 4cos(5x-2)

Solution:

Let y = 4cos(5x-t)

put 5x-2 = t. Then by differentiating both sides we get:

d/dx(5x-2) = dt

5dx = dt

dt/dx = 5.

y = 4 cost

dy/dx = (dy/dt)(dt/dx) = {d/dt{4cost)}{dt/dx)

dy/ dx = (4*-sint)(5).

We substitute t = (5x-2) and get:

Therfore,dy/dx = f'(x) = -20sin(5x-2)

For a function f(x) = g(h(x)), express h(x) as y.

Then f(x) = g(y), f’(x) = [d {g(y)}/ dy]*(dy/dx).

Here we have to find the derivative of f(x)= 4 cos (5x-2).

Let y=5x-2, this gives f(x)= 4 cos y

f’(x)= [d (4 cos y)/dy]*[d(5x-2)/dx]

We also know that the derivative of cos x= -sin x.

=> [d (4 cos y)/dy]= -4 sin y

[d(5x-2)/dx]= 5

Therefore f’(x)= [d (4 cos y)/dy]*[d(5x-2)/dx]

= (-4 sin y)*5

=-4*sin (5x-2)*5

=-20 sin (5x-2)

**Therefore the derivative of 4 cos (5x-2) is -20 sin (5x-2)**