Question

find dy/dx by implicit differentiation for: sinx+cosy = sinxcosy

Answer 1

what do you want to do with this

Answer 2

oh i didn't see find dy/dx sorry

Answer 3

I just edited it. No worries

Answer 4

ok so \[\frac{d}{dx} \sin(y)=\cos(y) \frac{dy}{dx} \text{ this is by chain rule }\]

Answer 5

I can differentiate the left side but not the right side.

Answer 6

I took the derivative of the outside (while leaving inside the same) and then times the derivative of the inside

Answer 7

the right side is a product

Answer 8

you must use product rule

Answer 9

\[\frac{d}{dx}uv=v \frac{d}{dx}u+u \frac{d}{dx} v \\ \frac{d}{dx}\sin(x)\cos(y)=\sin(x) \frac{d}{dx}(\cos(y))+\cos(y) \frac{d}{dx}(\sin(x))\]

Answer 10

so you need to differentiate cos(y) w.r.t x
and sin(x) w.r.t x

Answer 11

I understand the product rule. I come up with \[\sin (x)(-\sin y) y' + \cos(y)\cos(x)\]

Answer 12

Oh, this is starting to make a little more sense I think. Give me a minute.

Answer 13

you did awesome

Answer 14

Thanks. So let me get this straight: The reason I have a y' after the \[\sin(x)(-\sin y) \] part but not one after the [cos(y)] is because I'm taking the derivative for the first "y" but not the second?

Answer 15

yep

Answer 16

you only put y' when differentiating y

Answer 17

and in that second adden there was no differentiating y

Answer 18

addend *

Answer 19

Okay, thanks for clearing that up.

Answer 20

Anymore questions?

Answer 21

Yes, one more

Answer 22

When I move the y' to one side of the equation, I'm not sure how to simplify. I'm getting \[\sin(x)(-\sin y)y' + \sin(y)y' = \cos(x) -\cos(y)\cos(x)\]Is that okay so far or am i screwing something up?

Answer 23

\[\cos(x)-\sin(y) y'=\sin(x)(-\sin(y))y'+\cos(y)\cos(x) \\ \text{ get all y' terms on one side and all non y' terms on the other side } \\ \text{ add } \sin(y)y' \text{ and subtract }\cos(y)\cos(x) \text{ on both sides } \\ \cos(x)-\cos(y)\cos(x)=\sin(y)y'-\sin(x)\sin(y)y'\]
so that looks good

Answer 24

Now we did that so we could factor that y' out on the right hand side (and your case left hand side)

Answer 25

then divide by whatever y' was being multiplied by on both sides

Answer 26

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

Answer 27

I must go. But if you have more questions on this question @ganeshie8 or @campbell_st I think both have the expertise in the cal area.

Answer 28

Okay, thank you very much for all of your help! Greatly appreciated!

Answer 29

Np. Have fun. :)

Answer 30

@freckles, I finally got the answer! I realized I wasn't simplifying the right way. The final answer was \[\frac{\cos x(\cos y - 1)}{\sin y(\sin x -1)} \]

Answer 31

nice :) also you could simplify the given equation itself a bit before differentiating :

sinx+cosy = sinxcosy

sinx = cosy(sinx - 1)

sinx/(sinx-1) = cosy

Answer 32

you should get the final answer as : \[\large \dfrac{\cos x}{\sin y (\sin x -1)^2} = \dfrac{dy}{dx}\]

Answer 33

Thanks for the alternate solution @ganeshie8!