Differentiate a composite function

sbayla31

New member
Joined
Mar 6, 2012
Messages
4
Hello,

I am being asked to find the derivative of y = f(2f(2x)).

I know I should the chain rule for composite functions and the product rule, but I am getting stuck because of the constants thrown in there.

Thanks in advance for the help :)

(Just FYI, my history of Calculus knowledge is about 5 weeks of my high school Calculus course)
 
Hello,

I am being asked to find the derivative of y = f(2f(2x)).

I know I should the chain rule for composite functions and the product rule, but I am getting stuck because of the constants thrown in there.

Thanks in advance for the help :)

(Just FYI, my history of Calculus knowledge is about 5 weeks of my high school Calculus course)

These problems are best understood through substitution.

y = f(2f(2x)).

Let

u = 2x

u'= 2

v = 2f(2x) = 2*f(u)

dv/dx = 2 * df(u)/du * du/dx = 4 * df(u)/du

y = f(v)

dy/dx = df(v)/dv * dv/dx = 4 * df(u)/du * df(v)/dv = 4 * [f']2
 
I know I should the chain rule for composite functions and the product rule, but I am getting stuck because of the constants thrown in there.
I get a different answer from reply #2.

I read the question as \(\displaystyle D_x\left[f(2f(2x))\right]=~?\)

Let \(\displaystyle u=2x,~v=2f(u),~~w=f(v)\)

Then we want \(\displaystyle \displaystyle\dfrac{dw}{dv}\cdot\dfrac{dv}{du}\cdot\dfrac{du}{dx}\)

That gives \(\displaystyle 4f'(2f(2x))f'(2x)\)

Wolfram agrees.
 
Last edited:
Hello,

I am being asked to find the derivative of y = f(2f(2x)).

I read posts #2 and #3. It should be possible to see if either of

the answers could be eliminated by using an appropriate example.


Example:

Let f(x) = ax + b, with a and b belonging to the Real numbers, and a not equal to 0



Starting with:


y = f(2f(2x))


f(2x) = a(2x) + b

= 2ax + b



Then y =


f(2(2ax + b)) =


f(4ax + 2b) =


a(4ax + 2b) + b =


\(\displaystyle 4a^2x + 2ab + b\)


The derivative of this is \(\displaystyle 4a^2.\)


Now, Subhotosh Khan and pka can compare this result to their results.


And, if they agree on that one, then the test is inconclusive.



If the test is inconclusive, then ramp it up to another function,

say \(\displaystyle f(x) = ax^2 + bx + c, \\) to test the results of posts #2 and #3.


That'd not be a guarantee, but it would increase the chances of eliminating
one of the results.
 
Thanks for your answers.

Subhotosh Khan, I think I sort of understand your method, but I haven't learned things such as " u' " and "df(u)/du", and I'm not allowed to include such things in my answer.

pka, I don't know what the "Dx" bit is from this:


I read the question as \(\displaystyle D_x\left[f(2f(2x))\right]=~?\)

But I do like the idea of making substitutions.
I talked with some of my classmates today and we came up with this idea:

For y = f(2f(2x))

Let g(x) = 2f(2x)
so y = f(g(x))

Using the chain rule
dy/dx = f'(g(x)) * g'(x)

But what would g'(x) be? I tried the Wolfram website and I haven't learned "d/dx", so I'm really confused :(
*EDIT: I have recently been informed that the reason I am so confused is because I only know how to do "Explicit" differentiation. If anyone could differentiate this problem "explicitly", that would be great!

My friend says it would be 2*f'(2x) but I'm not totally sold.

Also, after substituting g'(x) in, I wouldn't have to go any further.
 
Last edited:
pka, I don't know what the "Dx" bit is from this:
But I do like the idea of making substitutions.
I talked with some of my classmates today and we came up with this idea:
\(\displaystyle D_x(f(x)=f'(x)\) is just notation for derivative with respect to \(\displaystyle x\)

Did you follow this link?

C
lick on the "show steps" tab.
You will see the chain of derivatives.
 
Yes, I did click "Show steps" but I didn't understand because I have not been taught "d/dx" or "df(x)/dx".
For my assignment, I have to differentiate it using explicit differentiation, and get dy/dx = (whatever it is).

I'm sorry, I really don't know much about calculus :/
 
Yes, I did click "Show steps" but I didn't understand because I have not been taught "d/dx" or "df(x)/dx". For my assignment, I have to differentiate it using explicit differentiation, and get dy/dx = (whatever it is).
That is truly an odd statement. \(\displaystyle \dfrac{df}{dx}\) is explicit differentiation.
 
pka, I'm sorry for confusing you. I was referring to d/dx, if that helps.

I would really, really appreciate if anyone could try to solve this problem the following way:

For y = f(2f(2x))

Let g(x) = 2f(2x)
so y = f(g(x))

Using the chain rule
dy/dx = f'(g(x)) * g'(x)

But what would g'(x) be? (without using d/dx)

My friend says it would be 2*f'(2x) but I'm not totally sold.

Also, after substituting g'(x) in, I wouldn't have to go any further.
 
pka, I'm sorry for confusing you. I was referring to d/dx, if that helps.
I would really, really appreciate if anyone could try to solve this problem the following way:
The only person you are confusing is yourself.
Let's take an example: \(\displaystyle y=\sin(2\sin(2x)).\)
If you do know, the derivative of \(\displaystyle \sin(x)\) is \(\displaystyle \cos(x).\)

So we get \(\displaystyle y'=\cos(2\sin(2x))(2)[\cos(2x)(2)]\)

Once again see this done.
 
Top