Question

cos(x+y)=ysin(x)

implicit differentiation, please help :)

Answer 1

the first derivative of the left side is:
\[\Large - \sin \left( {x + y} \right)\left( {1 + y'} \right)\]
since we have to apply the chain rule

Answer 2

and the first derivative of (x+y), is:
\[\Large {1 + y'}\]

Answer 3

\[y'\sin(x).\cos(x).1\]

Answer 4

for the other side?

Answer 5

in order to differentiate the right side, we have to apply the product rule, so we get:
\[\Large y'\sin x + y\cos x\]

Answer 6

OMG these do my head in LOL

Answer 7

just when you think you get the hang of it :(

Answer 8

now, is it ok?

Answer 9

yes I understand it :)

Answer 10

ok! so, after the differentiation of both sides, of your equation, we can write:
\[\Large - \sin \left( {x + y} \right)\left( {1 + y'} \right) = y'\sin x + y\cos x\]

Answer 11

yes I understand that, so would i then subtract ycosx from both sides?

Answer 12

we have to develop the left side, first. In order to that, we have to apply the distributive property of multiplication over addition, so we get:
\[ \Large - \sin \left( {x + y} \right) - y'\sin \left( {x + y} \right) = y'\sin x + y\cos x\]

Answer 13

yes ok I have that :)

Answer 14

ok! Now I add to both sides, this quantity:
\[\Large y'\sin \left( {x + y} \right)\]
so we can write:
\[ \Large - \sin \left( {x + y} \right) = y'\sin x + y\cos x + y'\sin \left( {x + y} \right)\]

Answer 15

then subtract ycosx ??

Answer 16

that's right!
so what do you get?

Answer 17

-sin(x+y)-ycos(x)=y'sin(x)(y)
??

Answer 18

I got this result:
\[ \Large - \sin \left( {x + y} \right) - y\cos x = y'\sin x + y'\sin \left( {x + y} \right)\]

Answer 19

oh sorry yes I went further and took out y'sin(x) as a common factor but that is not correct :(

Answer 20

ok! I guessed your reasoning. Now we can factor out y' at the right side, then we can write this:
\[\Large - \left[ {\sin \left( {x + y} \right) + y\cos x} \right] = y'\left[ {\sin x + \sin \left( {x + y} \right)} \right]\]

Answer 21

please note that I have factored out -1 at the left side

Answer 22

yes I got that :)

so can the bracket on the right become sin(2x+y)?

Answer 23

no, in order to simplify the bracket at the right side, we can apply the Prosthaphaeresis formulas

Answer 24

in order to find the expression for y', we have to divide both sides of last expression by this quantity:
\[\Large {\sin x + \sin \left( {x + y} \right)}\]
wha do you get?

Answer 25

\[\frac{ \sin(x+y)+ycosx }{ sinx+\sin(x+y) }=y'\]

Answer 26

that's right!

Answer 27

oops with a - in front

Answer 28

sorry you are right!

Answer 29

\[\Large y' = - \frac{{\sin \left( {x + y} \right) + y\cos x}}{{\sin x + \sin \left( {x + y} \right)}}\]

Answer 30

another?

Answer 31

yes!