Question

need help on the attachment

Answer 1

want to grind it out or do it a slightly simpler way?

Answer 2

is the simpler way slightly confusing?

Answer 3

no not really. it is just simple. ignore the "sine" and find the derivative of
\[\frac{x-4}{x+2}\] that is pretty easy

Answer 4

well my professor wants us to know trig identity, so I think its best to grind it out

Answer 5

using the "quotient rule" you get
\[\frac{(x+2)\times 1-(x+4)\times 1}{(x+2)^2}\]
\[\frac{6}{(x+2)^2}\]

Answer 6

then replace x by sine, and multiply by the derivative of sine, which is cosine, to get
\[\frac{6\cos(x)}{(\sin(x)+2)^2}\]

Answer 7

Oh,ok that is quick

Answer 8

otherwise it is still the quotient rule, just longer

Answer 9

you forget to write 6x on the numerator

Answer 10

\[\frac{(\sin(x)+2)\cos(x)-(\sin(x)+4)\cos(x)}{(\sin(x)+2)^2}\]

Answer 11

how did you just link 6 cos(X) when 6 was just by itself on the top

Answer 12

hold on. by the quotient rule you get
\[\frac{(x+2)\times 1-(x-4)\times 1}{(x+2)^2}=\frac{x+2-x+4}{(x+2)^2}=\frac{6}{(x+2)^2}\]

Answer 13

now you did not have x, you had sine
so i replace x by sine, and multiply my result by the derivative of sine which is cosine. that is why you get
\[\frac{6\cos(x)}{(\sin(x)+2)^2}\]

Answer 14

but if this is confusing or annoying forget it. we already have
\[\frac{(\sin(x)+2)\cos(x)-(\sin(x)-4)\cos(x)}{(\sin(x)+2)^2}\]

Answer 15

can you simplify that last step, please

Answer 16

the last one i wrote? this one?
\[\frac{(\sin(x)+2)\cos(x)-(\sin(x)-4)\cos(x)}{(\sin(x)+2)^2}\]?

Answer 17

yes

Answer 18

i am using
\[(\frac{f}{g})'=\frac{gf'-fg'}{g^2}\] with
\[f(x)=\sin(x)-4, f'(x)=\cos(x), g(x)=\sin(x)+2,g'(x)=\cos(x)\]

Answer 19

i did nothing at that step but plug the stuff in to the "quotient rule"

Answer 20

the hard part is going to be to turn the numerator in to 6 cos(x)

Answer 21

how do you turn the numerator into 6cos(x)= is there trig identity rule?