Differentiating f(2x+g(x)): Which is Correct?

[ggcheck]

Homework Statement

hi, I am supposed to differentiate the following:

f(2x+g(x))

I think the problem should be solved as follows:

f`(2x+g(x)) * (2x+g(x))` * g(x)`



but, my ta said to do the following:

f`(2x+g(x)) * (2x+g(x))`


please tell me, physicsforums, which is the correct way?

why?

 

[Avodyne]

Your TA is right.

Suppose you had f(h(x)). What would you say the derivative (with respect to x) is?

Got an answer?

OK, now let h(x)=2x+g(x).

 

[ggcheck]

"Suppose you had f(h(x)). What would you say the derivative (with respect to x) is?"

"OK, now let h(x)=2x+g(x)."

I am confused, but luckily I know where I am confused so I guess there is hope.

f`(h(x)) where h(x) = 2x+g(x)

wouldn't we have: df/dh * dh/dg * dg/dx ?

 

[Avodyne]

Let me try that again. Consider f(x^2). What is the derivative with respect to x?

 

[ggcheck]

f` * (x^2)`
?

 

[Avodyne]

OK, good. A clearer way to write this is f'(x^2)*(x^2)' And of course (x^2)'=2x.

Now consider a more general function h(x) as the argument of f: f(h(x)). The derivative of this with respect to x is f'(h(x))*h'(x).

Now, h(x) can be *any* function. It could be, for example, x^3. Or it could be x^3 + 2x. It doesn't matter what it is; we ALWAYS apply the formula f'(h(x))*h'(x).

So, if h(x)=g(x)+2x, we STILL apply the SAME formula. We don't multiply again by g'(x).

 

[ggcheck]

yeah, but I thought because it has another function inside of it (g(x)), we apply the chain rule again. sort of like we do with sin(cos(sinx)))

 

[Avodyne]

No, because g(x) is not inside of g(x)+2; that would mean something like f(g(g(x))+2x), with an extra "nesting".

 

[ggcheck]

with an extra "nesting".

this "nesting" is where I am confused? how exactly can I determine if there is a "nesting"?

 

[HallsofIvy]

this "nesting" is where I am confused? how exactly can I determine if there is a "nesting"?

yeah, but I thought because it has another function inside of it (g(x)), we apply the chain rule again. sort of like we do with sin(cos(sinx)))

This is an example of "nesting". The second sin(x) is "inside" (an argument of) the function cosine (and the whole assembly is a function of sine again).
[sin(cos(sin(x)))]' = cos(cos(sin(x)))[-sin(sin(x))][cos(x)]

If instead, you had sin(cos(x)+ sin(x)), the second sine is NOT "nested"- it is not an argument of the function cosine. The sum of the two functions is an argument of the sine function.
[sin(cos(x)+ sin(x))]'= cos(cos(x)+ sin(x))[-sin(x)+ cos(x)]

 

[ggcheck]

This is an example of "nesting". The second sin(x) is "inside" (an argument of) the function cosine (and the whole assembly is a function of sine again).
[sin(cos(sin(x)))]' = cos(cos(sin(x)))[-sin(sin(x))][cos(x)]

If instead, you had sin(cos(x)+ sin(x)), the second sine is NOT "nested"- it is not an argument of the function cosine. The sum of the two functions is an argument of the sine function.
[sin(cos(x)+ sin(x))]'= cos(cos(x)+ sin(x))[-sin(x)+ cos(x)]

but in : f(2x+g(x))

since g(x) is inside the argument of f(x) isn't g(x) nesting?

 

[HallsofIvy]

but in : f(2x+g(x))

since g(x) is inside the argument of f(x) isn't g(x) nesting?

It is not just g(x) but the whole 2x+g(x) that is inside f- you need to use the chain rule once, not twice: df(2x+g(x))/dx = (df/du)(du/dx) where u= 2x+ g(x). that is df/dx=f'(2x+g(x))(2+ g'(x)), exactly the same as the f`(2x+g(x)) * (2x+g(x))` your TA said- as long as you understand the " ' " to mean differentiation with respect to "the" variable- 2x+g(x) in f' and x in (2x+g(x))'. The point is that you have already used the chain rule in muliplying by (2x+ g(x))"- there is no reason to multiply by g'(x) again.

 

[ggcheck]

so we would only need to use the chain rule a second time if there was something "nesting" inside of g(x)?

 

[Avodyne]

Yes, if the argument of g(x) was something other than x, then we would need the chain rule a second time.

 

[ggcheck]

so it doesn't matter what g(x) is... it only matters what is in the argument of g(x)

g(x) could = (x^3+(x-1)^1/2 - 35x^14)^5/3 ... etc...

as long as only x is in the argument?

 

[Avodyne]

Yes!

 

[ggcheck]

Ok, so to sum things up... if we have:

f(2x+g(x))

and g(x)=(x^3+(x-1)^1/2 - 35x^14)^5/3 (copied from the last hypothetical)

then f`(2x+g(x))= f`(2x+g(x)) * (2x+g(x))` =
f`(2x+g(x)) * 5/3(x^3 +(x-1)1/2 - 35x^14)^2/3

 

[Avodyne]

The first line is correct. But then you did not compute (2x+g(x))' correctly.

(2x+g(x))' = 2 + g'(x), and

g(x) = h(x)^5/3, where h(x) = x^3+..., so

g'(x) = 5/3 h(x)^2/3 * h'(x)