Question

How do you find the derivative of `f(x)= cos (sin (4x))`?

Answer 1

`dy/dx=-4cos(4x)sin(sin(4x))`

Explanation:

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

`"given "y=f(g(h(x)))" then "`

`dy/dx=f'(g(h(x))xxg'(h(x))xxh'(x)`

`y=cos(sin(4x))`

`dy/dx=sin(sin(4x))xxd/dx(sin(4x))xxd/dx(4x)`

`color(white)(dy/dx)=sin(sin(4x))xx-cos(4x)xx4`

`color(white)(dy/dx)=-4cos(4x)sin(sin(4x))`

Answer 2

Rewrite `f(x)` as a composition of `3` functions, where we will use the chain rule to find the derivative of each function in a way that lets us find the final derivative:

`(df)/(dx) = -4 sin(sin(4x)) cos(4x)`

Explanation:

I'd break up the function composition into several parts. Here we have `cos(sin(4x))`, which does three things:

1. It multiplies the input by `4`,
2. It takes the sine value of the result from the above,
3. It takes the cosine value of the result from the above.

If we have `f(x) = cos(sin(4x))`, we could have

`f(x) = f_3(f_2(f_1(x)))` where

`f_1(x) = 4x`

`f_2(x) = sin(x)`

`f_3(x) = cos(x)`

Which should be doing each step from the inside, out. Then we can use the chain rule, which to me is more of a method than a rule. Here's how it goes:

Start from the innermost layer, taking the derivative of `f_1(x)` with respect to `x`.

`(df_1)/(dx) = d/dx f_1(x) = d/dx 4x = 4`

We also have:

`(df_1)/(dx) = 4 rarr df_1 = 4 dx`

What we did was solve for `df_1`, a tiny nudge in `f_1`, in terms of `dx`, a tiny nudge in `x`. Next, we'll take the derivative of `f_2(x)`, but in terms of `f_1(x)`:

`(df_2)/(df_1) = d/(df_1) f_2(f_1) = d/(df_1) sin(f_1) = cos(f_1)`

Solving for a tiny nudge in `df_2`, we "multiply" by `df_1`:

`(df_2)/(df_1) = cos(f_1) rarr df_2 = cos(f_1) df_1`

Then, we evaluate `f_1` and `df_1` in terms of `x` and `dx`:

`cos(f_1) df_1 = cos(4x) 4 dx = 4 cos(4x) dx`

Now, the tiny nudge is no longer in terms of `df_1`, but it is now in terms of `dx`:

`df_2 = 4 cos(4x) dx rarr (df_2)/(dx) = 4 cos(4x)`

And so is the derivative! What we have left to do is `f_3(x)`, whose tiny nudge we'll first find in terms of `df_2`:

`(df_3)/(df_2) = d/(df_2) f_3(f_2) = d/(df_2) cos(f_2) = -sin(f_2)`

`(df_3)/(df_2) = -sin(f_2) rarr df_3 = -sin(f_2) df_2`

Then we'll "unroll" things, first evaluating `f_2` in terms of `f_1` and `df_2` in terms of `dx`:

`df_3 = -sin(f_2) df_2 rarr -sin(sin(f_1)) 4 cos(4x) dx`

Then evaluating `f_1` in terms of `x`:

`rarr df_3 = -sin(sin(4x)) 4 cos(4x) dx`

Simplifying to make things look neater:

`df_3 = -4 sin(sin(4x)) cos(4x) dx`

And finally "divide" by `dx`:

`(df_3)/(dx) = -4 sin(sin(4x)) cos(4x)`

This is not only the derivative of `f_3(x)`, but because we had solved this through the derivatives of `f_2(x)` and `f_1(x)` from the inside out, this is our final answer, the derivative of `f(x) = cos(sin(4x))`:

`(df)/(dx) = -4 sin(sin(4x)) cos(4x)`