Question

Find the dy/dx of the following function.

y= 2cosx/ 1+sinx

PLEASE EXPLAIN.

Answer 1

product rule

Answer 2

Do you know the rule of division for differentiating.

Answer 3

forget division; product rule :)

Answer 4

I think division since its a fraction .

Answer 5

wouldnt it be the qoutient rule?

Answer 6

Im pretty sure you use the qoutient rule for solving, but I got lost can someone help?

Answer 7

y= 2cosx (1+sinx)^-1


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

Answer 8

yougot lost becasue your USING the quotient rule when the product rule is much much simpler to play with

Answer 9

sorry their divided by eachother not multiplied :/

Answer 10

actually, they are multiplied to each other ....

Answer 11

(2cosx)/(1+sinx) can i use the product rule if they are divided by eachother in the original eqaution?

Answer 12

\[\frac{2cos(x)}{(1+sin(x))}=\frac{2cos(x)}{1}*\frac{1}{(1+sin(x))}=2cos(x)*(1+sin(x))^{-1}\]

Answer 13

\[\frac{a}{b}=a*b^{-1}\]

Answer 14

\[y= 2cosx (1+sinx)^{-1}\]

\[y'= 2cosx' (1+sinx)^{-1} + 2cosx (1+sinx)'^{-1}\]

\[y' = -2sinx (1+sinx)^{-1} - 2cosx (1+sinx)^{-2} (cos(x))\]


\[y' = \left(-2sinx (1+sinx) - 2cosx (cos(x))\right) (1+sinx)^{-2}\]

\[y' = \frac{-2sinx (1+sinx) - 2cosx (cos(x))}{(1+sinx)^{2}}\]

Answer 15

thankyou i got to that point, the answer given is -2/1+sinx how do u get to that?

Answer 16

the answer "given" is a simplified version of this i beleive; and simplified versions pretty much depend on where you want to stop it at. if you plug this into a computer you get the answer back regardless of how simple youve made the final results :)

Answer 17

\[y' = \frac{-2sinx (1+sinx) - 2cosx (cos(x))}{(1+sinx)^{2}}\]

distribute thru and gather up all the likes

\[y' = \frac{-2sinx -2sin^2x - 2cos^2x}{(1+sinx)^{2}}\]

factor out the -2 and identify the s^2+c^2=1 id

\[y' = \frac{-2(sinx +sin^2x +cos^2x}{(1+sinx)^{2}}\]

\[y' = \frac{-2(sinx +1)}{(1+sinx)^{2}}\]

cancel out the like terms top to bottom

\[y' = \frac{-2}{1+sinx}\]
and tada!!