Recent content by silvershine

I'll definitely do that. Thank you!

So you would just be plugging 0 in for h and solving the limit normally?

Homework Statement A question we had for homework was: Which of the following is equivalent to limh->0 [arcsin((3(x+h))/4) – arcsin (3x/4)]/h ? Homework Equations There were multiple answer choices, but the correct answer is 3/(√16-9x^2).The Attempt at a Solution We've already walked through...

They have! Or they will, once I get more practice with them. Thanks for walking me through! I can do the rest of these bonus problems on my own now.

Alright. So to clarify: When you have a constant times a function, when you take the derivative, you don't have to use the product rule?

So then k = 8/5, right?

Sorry. So f'(x) would be 5k/(5x-3)?

Hm... So then would it be 5k/(5x-3)?

f'(x) would k[1/(5x+3)](5) + ln(5x-3), with the chain rule, right? Which would simplify to 5k/(5x-3) + ln(5x -3)?

It says ln[(5x-3)^k].

Sorry about posting it in the wrong section! This was a bonus question in my precalc homework. From the looks of it, the question says ln(5x - 3)^k. So would I do: 4 = kln(2) 4/k = ln(2) k = ln(2)/4?

Homework Statement Find the value of the constant, k, when f(x) = ln(5x-3)^k and f'(1) = 4Homework Equations f(x) = ln(5x-3)^k f'(1) = 4The Attempt at a Solution I've tried: f'(x) = [kln(5x-3)^(k-1)] [1/(5x-3)] (5) And when I simplified that, I got f'(x) = 5kln(5x-3)^(k-1) / (5x - 3) I tried...

This is a concept that confuses me. If you have a function where f(x) = ln(u)^k (where u is a function and k is an unknown constant), then how would you solve for the derivative? I've already guessed it would be f'(x) = [kln(u)^(k-1)](1/u)(u') Is that right?