Question

implicit differentiation: (e^y)(cosx)=1+sin(xy)

derivative of both sides:
(e^y (-sinx)+cosx(e^y))=sinxcosy+sinycosx
now what?

Answer 1

Are you trying to get dy/dx?

Answer 2

ya

Answer 3

Then y' needs to be showing up in some of those functions via chain rule.

Answer 4

how do i do that

Answer 5

If you differentiate y with respect to x, you assume y is a function of x, so you use chain rule and multiply by y'

Answer 6

e.g. \[\large \frac{d}{dx} e^y = y'e^y\]

Answer 7

i have no idea man lol

Answer 8

Also, sin(xy) is not the same as sin(x)sin(y), so you can't use the product rule like that.
You will have to use product rule while using chain rule on the xy in there.
It does get pretty messy...

Answer 9

oooh i didnt know that

Answer 10

Are you in Calc-I or Calc-II?
This is a pretty nasty one for a Calc-I course I would think.

Answer 11

calc 1

Answer 12

Yikes. Your professor is brutal.
Ok, let's step through this a little at a time.
Can you take the derivative of the left side using what I told you about using chain rule on y-as-a-function-of-x?

Answer 13

so well have
(e^y) d/dx(cos) + (cosx) d/dx(e^y)

Answer 14

Right.

Answer 15

chain rule for e^y equals 0(e)^(y-1) ??

Answer 16

No, d/dx of e^y = y'e^y.

Answer 17

It's the derivative of e^y, which is e^y, times the derivative of y, which is y'.

Answer 18

For chain rule: when you have a function-of-a-function, you multiply the derivatives of both functions together.

Answer 19

i didnt know that

Answer 20

d/dx of e^y = y'e^y. applies chain rule...

Answer 21

so well have
e^y(-sinx)+cosx(y'e^y)

Answer 22

Good, now for the right side, remember to use product rule on the xy, while doing chain rule on sin(xy).

Answer 23

(sinx) d/dx (siny)+ (siny) d/dx( sinx)

Answer 24

oops

Answer 25

tsk tsk, can't do that.

Answer 26

nm that well have

d/dx(1) + sin [x d/dx (y) + y d/dx (x)]

Answer 27

For chain rule, multiply derivative of sin by the derivative of xy, for the derivative of xy, use product rule.

Answer 28

That looks good, but don't forget to d/dx the sin itself too.

Answer 29

ya just noticed that

Answer 30

so we should have

(cosx)[(x)dy/dx + (y)(1)]

Answer 31

Ok, to summarize, you should have this by this point:
\[\large e^ycos(x)=1+sin(xy)\]
\[\large \frac{d}{dx} e^ycos(x) = e^y(-sin(x))+cos(x)y'e^y\]
\[\large \frac{d}{dx}( 1+sin(xy)) = 0+cos(xy)(xy'+y)\]

Answer 32

Cos(xy), not just cos(x)

Answer 33

Now set the left-side derivative equal to the right-side derivative and simplify.

Answer 34

e^y(-sinx)+cosx(y'e^y)=(cosxy) (x)dy/dx + (y)
-y
e^y(-sinx)+cosx(y'e^y) - y = cos(xy)dy/dx
divide cosxy
(e^y(-sinx)+cosx(y'e^y) - y)/cos(xy) = dy/dx

Answer 35

Ermmm, I don't think that works because you still have y' mixed up with the rest of the junk.

Answer 36

i want this problem to end already lol

Answer 37

Yeah, it's a beast!

The equation simplifies to
\[\large -(e^y)(sin(x))+(cos(x))(e^y)(y') = (x)(cos(xy))(y')+(y)cos(xy)\]
Then get all the groups of y' on one side and everything else on the other
\[\large (cos(x))(e^y)(y') - (x)(cos(xy))(y') = (y)cos(xy) + (e^y)(sin(x))\]

Answer 38

i see then we factor maybe?

Answer 39

Yes, you have to factor out y' so you can solve for it
\[\large [(cos(x))(e^y) - (x)(cos(xy))](y') = (y)cos(xy) + (e^y)(sin(x))\]

Answer 40

One last step to isolate the y' . .
\[\large y' = \frac{ycos(xy) + e^ysin(x)}{e^y cos(x) - xcos(xy)}\]

Answer 41

CliffSedge your a lifesaver thanks for showing me the steps

Answer 42

You're welcome. Tell your professor I called him, "brutal." This was some serious differentiation. It should all be things you know how to do, but it's a damn lot of it all together!
Whew!