Unbelievable confusion in calculating the derivative of a function

[IliaPreston]

Hello everyone.
Let's say we have a function as follows:

f(x) = 3x² - 2x + 7

The goal is to find f '(5x).
I try to calculate that in two different ways:

Method 1:

f '(x) = 6x - 2
f '(5x) = 6(5x) - 2
f '(5x) = 30x - 2

Method 2:

f(x) = 3x² - 2x + 7
f(5x) = 3(5x)² - 2(5x) + 7
f(5x) = 75x² - 10x + 7
f '(5x) = 150x - 10

That is two completely different results for f '(5x) !!!!!!

How can that be?
If you solve a problem in two (or many) different ways, the result should be the same.
So, why are the results different in this case?

Which one is correct?
And what is wrong in the line of reasoning that leads to the other?

Please help.
Thanks.

[OptionBase1]

Meh, what you've asked is basically the same question as to why the below produces different results:

Let f(x) = 3x^2 - 2x + 7
Calculate f'(2).

Method 1:
f(x) = 3x^2 - 2x + 7
f'(x) = 6x - 2
f'(2) = 6*2 - 2
f'(2) = 10

Method 2:
f(x) = 3x^2 - 2x + 7
f(2) = 3*(2^2) - 2*2 + 7
f(2) = 12 - 4 + 7
f(2) = 15
f'(2) = 0


Order matters.

[OptionBase1]

Here's another example that hopefully reinforces things:

Let f(x) = x
Find f'(sin(x))

Method 1:
f(x) = x
f'(x) = 1
f'(sin(x)) = 1

Method 2:
f(x) = x
f(sin(x)) = sin(x)
f'(sin(x)) = cos(x)


Which one do you think is correct?

Basically, the question posed is, what is the slope of f(x) at sin(x)? And obviously, since f(x) = x, the slope is 1 for any value given, regardless of how that value is arrived at, whether it is a constant or the result of evaluating an additional function of x.

[Thierry69]

I am not as good in math,
Just ask ChatGPT

[OptionBase1]

Or let a = 5x.

We can confidently say that:
f(a) = 3a^2 - 2a + 7
f'(a) = 6a - 2

Substitute 5x for a:
f'(5x) = 6*5x - 2
f'(5x) = 30x - 2