Part help: Calculus help

Iz-A-Demon

An Unknown Location

Calculus help

Okay, I need to differentiate this problem.

y = (ln tan x)^2

And the only thing that’s throwing me is the square (^2) part. Am I allowed to bring that down, or do I have to treat it like a chain rule thing?

This closed post was written 3 months, 2 weeks ago | V/U/S: 87, 11, 2 | Edit Post | Report Post

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Since writing this post Iz-A-Demon may have helped people, but has not within the last 4 days. Iz-A-Demon is a verified member, has been around for 5 months, 2 weeks and has 15 posts and 24 replies to their name.

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Calculus, Chain rule, Square, Chain, Rule, Part, differentiation, Throwing, tan (How Tags Affect Reciprocity)

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Belief 2 #
Roslindale, MA, US | 3 months, 2 weeks ago (2 minutes after post)

Yeah but remeber to differiate for whats in the paranthesis as well.

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Iz-A-Demon 4 #
An Unknown Location | 3 months, 2 weeks ago (9 minutes after post)

Yeah, I got that. Just wasn’t sure if I was on the right track or not. So, it should be something like
y = f(g(x))

u = g(x) = ln tan x
f(u) = u^2

y’ = f’(x) * g’(x)
thus
y’ = 2(ln tan x) * [ 1/(sec^2 x) ]

Or is that way off?

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Krieg-ha, Bandolo! 0 #
An Unknown Location | 3 months, 2 weeks ago (21 minutes after post)

Iz-A-Demon wrote:
Yeah, I got that. Just wasn’t sure if I was on the right track or not. So, it should be something like
y = f(g(x))

u = g(x) = ln tan x
f(u) = u^2

y’ = f’(x) * g’(x)
thus
y’ = 2(ln tan x) * [ 1/(sec^2 x) ]

Or is that way off?

just some mistakes in typing, but that’s right
(df/dg * dg/dx ), not (df/dx * dg/dx )

Krieg-ha, Bandolo! 0 #
An Unknown Location | 3 months, 2 weeks ago (23 minutes after post)

actually, df/du * du/dg * dg/dx, where f(u) = u^2, u(g) = ln(g), g(x) = tan(x)

Iz-A-Demon 4 #
An Unknown Location | 3 months, 2 weeks ago (27 minutes after post)

So, wait. I think I’m confused now. Is what I have right, just the wording is wrong, or is the answer part wrong?

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Krieg-ha, Bandolo! 0 #
An Unknown Location | 3 months, 2 weeks ago (28 minutes after post)

after my second post I dunno lol
didn’t try finishing =)

Krieg-ha, Bandolo! 0 #
An Unknown Location | 3 months, 2 weeks ago (30 minutes after post)

its…
2*u*(1/g)*D(tan(x)) = 2*ln(tan(x))/tan(x) * D(tan(x))
but I don’t remember (tan(x))’

Iz-A-Demon 4 #
An Unknown Location | 3 months, 2 weeks ago (34 minutes after post)

Oh, ok. I was taking the derivative of ‘ln tan x’ in the second not just ‘tan x’.

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Krieg-ha, Bandolo! 0 #
An Unknown Location | 3 months, 2 weeks ago (36 minutes after post)

yep

Iz-A-Demon 4 #
An Unknown Location | 3 months, 2 weeks ago (37 minutes after post)

Ok, thank you for your help.

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