# Find the derivative (1 Viewer)

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#### endeavor

Find f'(x) if it is known that
$$\frac{d}{dx}[f(2x)] = x^2$$

I let u(x) = 2x, then
$$\frac{d}{dx}[f(u)] = \frac{dy}{du} \frac{du}{dx}$$
$$\frac{d}{dx}[f(2x)] = 2 \frac{dy}{du}$$
therefore
$$\frac{dy}{du} = \frac{1}{2}x^2$$
then
$$f'(x) = \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$$
$$= (\frac{1}{2}x^2) (2)$$
$$= x^2$$
why doesn't this work??

#### A_I_

u have f'(2x) = x^2

first find f(2x) and try to write it as a function of (2x)

and then after u do that substitute x for 2x
you therefore find f(x) finally derive it and you obtain f'(x)

does it make any sense?

#### endeavor

yeah, i think so.

#### A_I_

let me know what your answer is to see if you did it right

#### endeavor

Well, i kinda cheated and looked at the solutions manual:tongue2:
Using the chain rule:
$$\frac{d}{dx}[f(2x)] = 2 f'(2x) = x^2$$
$$f'(2x) = \frac{1}{2}x^2$$
then let u(x) = 2x
$$f'(u) = \frac{1}{2} (\frac{u}{2})^2$$
$$f'(u) = \frac{1}{8} u^2$$
then substitute x for u:
$$f'(x) = \frac{1}{8} x^2$$

Is there an easy to understand explanation why my original method did not work?

#### A_I_

hehe good

my method was simple: f'(2x) = x^2 --> f(2x) = x^3 / 3

we multiply the denominator and numerator by 8 --> f(2x) = 8x^3 / 24

---> f(2x) = (2x)^3 / 24 ---> f(x) = x^3 / 24 --> f'(x) = x^2 / 8

#### A_I_

your mistake was taking partial derivatives

u write f'(x) = dy/dx = (dy/du)/(du/dx)

since y and u are both functions of x u can not apply the partial derivative (the chain rule formula)

#### endeavor

Thanks A_I_, I like your method :tongue:

can you expand on what you said my mistake was? I'm not sure I understand.

#### A_I_

the chain rule is used when u have:

f(x,y) and x=g(t) and y=r(t)

but since your f and your u are both functions of x,
thus you can not use the chain rule and you can not say:

f'(x) = (dy/du)*(du/dx)

ok?

#### endeavor

Yeah, i think I understand.

Can you explain that function syntax? I've seen it before, but I've always learned the simple f(x) not f(x,y)

and for composite functions (using the chain rule), f(x) = g(r(x))

#### A_I_

what do you exactly want to know?
do you want an example of the chain rule?

what you wrote: f(x)= g(r(x)) is (gor)(x) and is different from the chain rule.

do you have an im?

#### endeavor

i know what the chain rule is.
i just want to know what f(x,y) means and what "f(x,y) and x=g(t) and y=r(t)" means.

here you go

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#### VietDao29

Homework Helper
A_I_ said:
your mistake was taking partial derivatives

u write f'(x) = dy/dx = (dy/du)/(du/dx)

since y and u are both functions of x u can not apply the partial derivative (the chain rule formula)
What do you mean by partial derivative here??? You only encounter partial derivative in multi-variable function, not 1 variable function like f(x)!!!!
A_I_ said:
the chain rule is used when u have:

f(x,y) and x=g(t) and y=r(t)

but since your f and your u are both functions of x,
thus you can not use the chain rule and you can not say:

f'(x) = (dy/du)*(du/dx)

ok?
???
NOOO!!!!!! What do you mean by this??? I am TOTALLY lost!!!
Why can't you use the chain rule in that case? Am I misssing something?
Please look back at your cal textbook, see the part that covers the chain rule.
-------------------
$$\frac{df}{dx} = \frac{df}{du} \times \frac{du}{dx}$$
Example:
$$\frac{df(2x)}{dx} = \frac{df(2x)}{d(2x)} \times \frac{d(2x)}{dx} = 2 f'(2x)$$.
Now, let's do your problem in a slightly different way:
$$\frac{df(2x)}{dx} = 2 f'(2x) = x ^ 2$$
$$\Rightarrow f'(2x) = \frac{x ^ 2}{2}$$
Now, let y = 2x, we have:
$$f'(y) = \frac{(2x) ^ 2}{8} = \frac{y ^ 2}{8}$$
Now, what's f'(x)?
Can you go from here? Is there anything unclear? :)
-------------------
Whoops, looking back at some previous posts of this thread, I saw that you've looked through the manual.
The only error you made is that you were trying to find dy / dx, which means you were finding d(f(2x)) / dx, not d(f(x)) / dx (which means the same as f'(x)).
Is there anything unclear, endeavor? :)

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#### A_I_

VietDao29 said:
What do you mean by partial derivative here??? You only encounter partial derivative in multi-variable function, not 1 variable function like f(x)!!!!

???
NOOO!!!!!! What do you mean by this??? I am TOTALLY lost!!!
Why can't you use the chain rule in that case? Am I misssing something?
Please look back at your cal textbook, see the part that covers the chain rule.
-------------------
$$\frac{df}{dx} = \frac{df}{du} \times \frac{du}{dx}$$
Example:
$$\frac{df(2x)}{dx} = \frac{df(2x)}{d(2x)} \times \frac{d(2x)}{dx} = 2 f'(2x)$$.
Now, let's do your problem in a slightly different way:
$$\frac{df(2x)}{dx} = 2 f'(2x) = x ^ 2$$
$$\Rightarrow f'(2x) = \frac{x ^ 2}{2}$$
Now, let y = 2x, we have:
$$f'(y) = \frac{(2x) ^ 2}{8} = \frac{y ^ 2}{8}$$
Now, what's f'(x)?
Can you go from here? Is there anything unclear? :)
for the first part, thats what i was trying to tell him,
that we only use the chain rule when we have a multivariable function, which is not the case.
As for the second part, i am not sure about it, because i know you can only use the chain rule with multivariable function.
Did u consider x and 2x to be two different variables, if yes, then it works.

#### VietDao29

Homework Helper
A_I_ said:
for the first part, thats what i was trying to tell him,
that we only use the chain rule when we have a multivariable function, which is not the case.
As for the second part, i am not sure about it, because i know you can only use the chain rule with multivariable function.
Did u consider x and 2x to be two different variables, if yes, then it works.
What do you mean???
Chain rule can be used for both multivariable functions, and 1 variable function.
You must have studied $$\frac{df}{dx} = \frac{df}{du} \times \frac{du}{dx}$$ BEFORE studying multi-variable functions, no?
As I told you before, you should re-read your calculus textbook, just look up the chapter for Chain rule (or you can just click on the link) for one variable function. That won't do you any harm, I promise.

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#### A_I_

I get what you've said.
Thanks for the explanation

#### endeavor

So when in my first post I wrote:
$$\frac{dy}{du}$$
that really means
$$\frac{d f(2x)}{du}$$
since y = f(2x) ?

and then later on, when I wrote:
$$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$$
I was actually finding f'(2x), which was already given??

#### VietDao29

Homework Helper
endeavor said:
So when in my first post I wrote:
$$\frac{dy}{du}$$
that really means
$$\frac{d f(2x)}{du}$$
since y = f(2x) ?
Yes, this is correct. :)

and then later on, when I wrote:
$$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$$
I was actually finding f'(2x), which was already given??
Nah, this is not correct, you are finding:
$$\frac{df(2x)}{dx}$$, not f'(2x).
$$\frac{df(2x)}{dx} \neq f'(2x)$$.
To find f'(2x), you must find: $$\frac{df(2x)}{d(2x)}$$.
Can you get this? :)

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