Question

Is this correct?
Derivative of y = sqrt[(x^2+1)(sin3(x)]

Answer 1

y' = (x^2+1)^1/2(3/2)(sinx)(cosx)+(sinx)^3/2 (1/2)(x^2+1)^-1/2 (2x)

Answer 2

im confused haha sorry

Answer 3

Hmm I can't seem to figure out what you did bobo :( it's kind of hard to read.

Answer 4

Can you teach me the right way?

Answer 5

\[\Large y=\left[(x^2+1)\sin3x\right]^{1/2}\]
Applying Power Rule first,\[\Large y'=\frac{1}{2}\left[(x^2+1)\sin3x\right]^{-1/2}\color{royalblue}{\left[(x^2+1)\sin3x\right]'}\]
Chain Rule tells us that we need to take a copy of the inner function and multiply by its derivative. So we still need to take the derivative of this blue portion.
Understand the power rule step?

Answer 6

yes

Answer 7

Then looks like we need to apply the product rule to the blue part, giving us this setup:\[\large y'=\frac{1}{2}\left[(x^2+1)\sin3x\right]^{-1/2}\left[\color{#CC0033}{(x^2+1)'}\sin3x+(x^2+1)\color{#CC0033}{(\sin3x)'}\right]\]

Answer 8

Okay I follow that

Answer 9

From here, you need to take the derivative of the red parts.
What do you get for those? :x

Answer 10

2x and cos3x?

Answer 11

Hmm close! Don't forget the chain rule on your sin3x. It should give us an extra 3, right?\[\large y'=\frac{1}{2}\left[(x^2+1)\sin3x\right]^{-1/2}\left[\color{royalblue}{(2x)}\sin3x+(x^2+1)\color{royalblue}{(3\cos3x)}\right]\]

Answer 12

okk

Answer 13

is there not something that comes after?

Answer 14

Mmmm no nothing else \c:/

Answer 15

Thank you very much @zepdrix
You're a great teacher

Answer 16

yay team :D