Question

I need help with this one derivative of y=
equation at the bottom

Answer 1

\[Y=\sin \left[ \cos \left( 2x-4 \right) \right]\]
\[y \prime=\cos \left[ \cos \left( 2x-4 \right) \right]\left\{ -\sin \left( 2x-4 \right) \right \} \times2\]
\[y \prime=-2\cos \left[ \cos \left( 2x-4 \right) \right]\sin \left( 2x-4 \right)\]

Answer 2

first you have to take differential of sin only,then cos and finally 2x-4

Answer 3

ooh thank you i understand now:)

Answer 4

yw

Answer 5

I need help wit this one derivative of y= \[\frac{ (2x+3)^{3} }{ (x+1)^{3} }\]

Answer 6

So ummm Quotient Rule, yes?

Answer 7

yes and chain rule too

Answer 8

\[\Large y'\quad=\quad \frac{\color{royalblue}{\left[(2x+3)^3\right]'}(x+1)^3-(2x+3)^3\color{royalblue}{\left[(x+1)^3\right]'}}{\left[(x+1)^3\right]^2}\]

Answer 9

the answer is y'= \[\frac{ -3(2x+3)^{2} }{ (x+1)^{4} }\] but idk how

Answer 10

Ok there is our quotient rule setup, so we need to take the derivative of the blue parts.

Answer 11

What do you get for the first blue part? :oU

Answer 12

omg nvm i figured it out how to do it!:)