Question

How do you implicitly differentiate (sinx+cosy)^2=1?

Answer 1

where are you stuck?

Answer 2

I tried using the chain rule for it, but my answer doesnt simplify at all :s

Answer 3

why don't you show us what you tried.

Answer 4

2(sinx+cosy)*(cosx-siny y') = 0

I multiplied them together and got

y' = -2cosxcosy - 2sinxcosx/ -2(sinxsiny - cosysiny)

so from there im stuck

Answer 5

\[2(\sin(x)+\cos(y))*(\cos(x)-\sin(y) y') = 0 \]

\[(\sin(x)+\cos(y))\cos(x)-(\sin(x)+\cos(y))\sin(y) y' = 0 \]
\[(\sin(x)+\cos(y))\cos(x)=(\sin(x)+\cos(y))\sin(y) y' \]

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

\[y'=\frac{\cos(x)}{\sin(y)}\]

Answer 6

he divided by 2 (which drops out because t's all =0)
then distributed sinx+cosy to cosx and -sinx in the next set of parentheses

Answer 7

he divided by 2 (which drops out because t's all =0)
then distributed (sinx+cosy) to cosx and -sinx in the next set of parentheses

Answer 8

okay i kinda get that, but when you distribute the (sinx+cosy) to cosx and -sinx to the next would you have to foil them out? i end up getting

sinxcosx -sinysinx + cosxcosy -siny y' cosy = 0

Answer 9

to make it more clear:
let\[\sin x+\cos y=a\]we then have\[a(\cos x-\sin yy')\]distributing we get\[a\cos x-a\sin yy'\]subbing back in \(\sin x+\cos y=a\) we get Zarkon's answer

Answer 10

oh okay, so he didnt completely distribute them out but kept in a(cosx - sinyy') form
i think i get it now, thanks !!

Answer 11

Yeah, Zarkon is not so much about explanation, but he will answer the most difficult question with ease. He leaves the groundlings like me to explain his techniques.

Answer 12

haha yeah, thanks for the explanation though because i find that more important than just the answer!

Answer 13

that's my philosophy as well, so I'm happy to oblige :)