Help derivation arctan

[hellothere]

hi guys (and gals)

i have a math presentation on monday and i really cant solve this derivate. could you help me?

f(x)= arctan^2 (ln (x-1))

f'(x)= ?

thanks in advance

 

[Deleted member 4993]

hi guys (and gals)

i have a math presentation on monday and i really cant solve this derivate. could you help me?

f(x)= arctan^2 (ln (x-1))

I assume your problem is:

f(x) = [arctan{ln(x-1)}]²

f'(x)= ?

thanks in advance

Use substitution

u = ln(x-1)

f = [tan^-1(u)]²

then

\(\displaystyle \frac{df}{dx} \ = \ \frac{df}{du} \cdot \frac{du}{dx}\)

 

[hellothere]

im sorry to bother you again.

but im from Portugal and in here we really dont use that method and i dont quite fully understand it.

could you break it down step by step please?

 

[HallsofIvy]

Then what do you know?

Do you know the derivative of arctan(x)?
Do you know the derivative of ln(x)?
Do you know the "chain rule"?

 

[hellothere]

of course i know the chain rule.

and those other derivatives tou mentioned. i just dont understand that thing with the dx´s and du´s

is that just another way of representing the chain rule?

 

[Deleted member 4993]

of course i know the chain rule.

and those other derivatives tou mentioned. i just dont understand that thing with the dx´s and du´s

is that just another way of representing the chain rule?

If by - that - you mean:

\(\displaystyle \dfrac{df}{dx} \ = \ \dfrac{df}{du} \cdot \dfrac{du}{dx}\)


Then yes - it is the chain rule mathematically expressed (everywhere in the world using english script).

 

[HallsofIvy]

of course i know the chain rule.

and those other derivatives tou mentioned. i just dont understand that thing with the dx´s and du´s

is that just another way of representing the chain rule?

I'm not sure I would call it another way! Exactly what form did you learn for the chain rule?

In any case, the derivative of f(x)= arctan^2 (ln (x-1)) is 2 times arctan(ln(x-1)) times the derivative of arctan(x)(replacing "x" in the derivative with "ln(x- 1)") times the derivative of ln(x) (replacing "x" in the derivative with "x- 1") times the derivative of x- 1. You say you know all those derivatives, so- go to it.

A very precise statement of the chain rule for this would be:
Let u= x- 1 so f(u)= arctan^2(ln(u)).
Let v= ln(u) so f(v)= arctan^2(v).
Let w= arctan(v) so f(w)= w^2.

Then the chain rule says that \(\displaystyle \frac{df}{dx}= \frac{df}{dw}\frac{dw}{dv}\frac{dv}{du}\frac{du}{dx}\).