# Calc Implicit Differentiation Problem?

• Jan 1st 2010, 06:44 PM
Milou
Calc Implicit Differentiation Problem?
Okay, so I'm trying to do my Calc homework and for some reason, I'm getting stuck on a differentiation problem.

So, it's basically (sin(x)/(x))+(sin(y)/(y))=C, where C is some constant. Obviously the constant would disappear, so I get: (after simplifying)
((xcos(x)-sin(x))/(x^2))+((y'cos(y')-sin(y'))/(y'^2))=0

My problem is getting a y' on one side of the equation and the rest on the other side so that I can solve it. (Provided I actually differentiated it correctly above...)

I do have one point on the curve, (pi/3, pi/4) if that helps at all, but I need to get it to the point where I can plug in a value for x and get the y value, and I'm kinda stuck.

I would really appreciate any help. :-)

Thank you!
• Jan 1st 2010, 06:54 PM
eXist
Be careful when differentiating, you have:

$\frac{xcos(x)-sin(x)}{x^2}+\frac{y'cos(y')-sin(y')}{y^2}=0$

When it should be:

$\frac{xcos(x)-sin(x)}{x^2}+\frac{y'cos(y)y-sin(y)y'}{y^2}=0$

Now, solve the about for $y'$ by moving everything with it to one side, then factoring it out, then divide by what remains.

This will get you an equation for $y'$ in terms of $y$ and $x$, and then use the (x,y) point you have given to find $y'$ at that point.
• Jan 1st 2010, 07:37 PM
Milou
Thanks!
Okay.... I think I get it. Thanks!!! :D

Edit:

Ahh - so I forgot about the inside / outside thing when differentiating.
• Jan 1st 2010, 07:57 PM
Soroban
Hello, Milou!

Quote:

Given: . $\frac{\sin x}{x} + \frac{\sin y}{y} \:=\:C$ . [1]

Find $y'.$

First, eliminate the fractions . . .

Multiply by $xy: \;\;y\sin x + x\sin y \:=\:Cxy$

Differentiate implicitly: . $y\cos x + y'\sin x + x\cos y\cdot y' + \sin y \;=\;Cxy' + Cy$

Re-arrange terms: . $y'\sin x + y'x\cos y - Cxy' \;=\;Cy - y\cos x - \sin y$

Factor: . $y'(\sin x + x\cos y - Cx) \;=\;Cy - y\cos x - \sin y$

Therefore: . $y' \;=\;\frac{Cy - y\cos x - \sin y}{\sin x + x\cos y - Cx}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

This answer looks different from yours, which has no $C.$

From [1], we have: . $C \;=\;\frac{y\sin x + x\sin y}{xy}$

Substitute this into my answer and simplify.

And we get: . $y' \;=\;\frac{y^2(\sin x - x\cos x)}{x^2(y\cos y - \sin y)}$

This is answer you should get, using the Quotient Rule.

• Jan 1st 2010, 08:05 PM
Milou
Thank you!

It's so nice of you to help me.

I like your method - fractions confuse me, so that really helps. :) I'll try it for myself right now.

Thanks!!!