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

2 Answers
Dec 19, 2017

Answer:

#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

Answer:

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)#