Question

find the first derivative of y=sin[(5t+6)^-1/9]

Answer 1

\[\Large y\quad=\quad \sin\left[(5t+6)^{-1/9}\right]\]

Answer 2

So we take the derivative of the outermost function (sine in this case), which gives us cosine.
\[\Large y'\quad=\quad \cos\left[(5t+6)^{-1/9}\right]\color{royalblue}{\left[(5t+6)^{-1/9}\right]'}\]
Then the chain rule tells us to make a copy of the inside and take it's derivative.

Answer 3

okay so do I go:

cos[(5t+6)^-1/9][-1/9(5t+6)^-10/9]

Answer 4

@zepdrix How do I continue to expand it?
Also is the derivative of [(5t+6)^-1/9)] = [-1/9(5t+6)^-10/9]

Answer 5

Your derivative of the blue part looks correct, now we have to apply the chain rule once again.\[\Large y'\quad=\quad \cos\left[(5t+6)^{-1/9}\right]\color{royalblue}{\left[-\frac{1}{9}(5t+6)^{-10/9}\right]}\color{green}{\left(5t+6\right)'}\]

Answer 6

cos[(5t+6)^-1/9][-1/9(5t+6)^10/9][5] is this correct? @zepdrix

Answer 7

Yah that looks good :)
Multiplication is `commutative` (we can multiply in any order) , so we can rearrange stuff to make it look a little nicer.

Answer 8

So, how how would I simplify the equation so it looks "prettier"

Answer 9

I would probably bring the "stuff" in front of the cosine, so it's clearer what exactly is the argument of our trig function.\[\Large y'\quad=\quad -\frac{5}{9}(5t+6)^{-10/9}\cdot\cos\left[(5t+6)^{1/9}\right]\]

Answer 10

Would this be simplified as well:

\[\frac{ -5\cos(\frac{ 1 }{ (5t+6)^{-1/9} }) }{ 9(5t+6)^{10/9} }\]

Answer 11

@zepdrix

Answer 12

@zepdrix Thank you so much for your help