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

2 Answers
Dec 19, 2017

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))

Dec 19, 2017

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)