Question about Implicit Differentiation

**FalconPUNCH!** · October 23rd 2007, 06:12 PM

My question is when do you multiply by y' or $\frac d{dx}$ when trying to get the derivative?

I know that in this problem $x^3 + y^3 = 25$ that you can do it $3x + 3y (y') = 0$

In more complicated equations I don't know when to use it. For example in this problem

$sin(x) + cos(y) = sin(x)cos(y)$

I don't know where y' belongs. Can anyone help me find out when I have to multiply by y'. I know you are supposed to when you're using the chain rule but I don't know how to do it in that type of problem.

-FP!

**sixstringartist** · October 23rd 2007, 06:21 PM

You're not simply multiplying by y'. y' is a component in the chain rule due to the implicit nature of the problem.

I believe your first problem is incorrect.

$x^3 + y^3 = 25 $

$ $ $3x^2 + 3y^2 (y') = 0 $

Here you are taking the derivative of every term. Because this is implicit, the derivative of y^3 is 3(y)^2 * (y').
Likewise if your problem was:
$ x^3 + \sin{y^2} = 25 $
than,
$ 3x^2 + \cos{y^2} * \frac {d(y^2)}{dy} = 0$
[/color]

**FalconPUNCH!** · October 23rd 2007, 06:30 PM

Originally Posted by sixstringartist

You're not simply multiplying by y'. y' is a component in the chain rule due to the implicit nature of the problem.

I believe your first problem is incorrect.

$x^3 + y^3 = 25 $

$3x^2 + 3y^2 (y') = 0 $

Here you are taking the derivative of every term. Because this is implicit, the derivative of y^3 is 3(y)^2 * (y').
Likewise if your problem was:
$ x^3 + \sin{y^2} = 25 $
than,
$ 3x^2 + \cos{y^2} * \frac {d(y^2)}{dy} = 0$
[/color]

Yeah I forgot the to square each term. There are problems that I have trouble doing this in though. Problems like the one I mentioned in my OP $sin(x) + cos(y) = sin(x)cos(y)$

$cos(x) + (-sin(y)) y' = cos(x)cos(y) + sin(x)(-sin(y)) y'$

$(-sin(y)) y' - sin(x)(-sin(y)) y' = cos(x)cos(y) - cos(x)$

$y' (-sin(y) - sin(x)(-sin(y)) = cos(x)cos(y) - cos(x)$

$y' = \frac {cos(x)cos(y) - cos(x)}{-sin(y) - sin(x)(-sin(y))}$

Is that the correct way to use it?

**sixstringartist** · October 23rd 2007, 06:57 PM

Yes that appears to be correct. I looked over your math, but I didnt write it out myself so I may have overlooked an algebraic error but the method of implicit differentiation appears correct.

**FalconPUNCH!** · October 23rd 2007, 07:07 PM

Originally Posted by sixstringartist

Yes that appears to be correct. I looked over your math, but I didnt write it out myself so I may have overlooked an algebraic error but the method of implicit differentiation appears correct.

Ok thank you for your help. I understand it much more clear now.

**ticbol** · October 23rd 2007, 09:34 PM

Originally Posted by FalconPUNCH!

My question is when do you multiply by y' or $\frac d{dx}$ when trying to get the derivative?

I know that in this problem $x^3 + y^3 = 25$ that you can do it $3x + 3y (y') = 0$

In more complicated equations I don't know when to use it. For example in this problem

$sin(x) + cos(y) = sin(x)cos(y)$

I don't know where y' belongs. Can anyone help me find out when I have to multiply by y'. I know you are supposed to when you're using the chain rule but I don't know how to do it in that type of problem.

-FP!

Here is one way.

Remember with respect to what variable you are differentiating.
If it is with respect to x, then all differentiated variables will end up with dx as the denominator. (Even me, I'm lost in that explanation.

] In actual, it is like this:
x^3 +y^3 = 25
Differentiate both sides with respect to x,
(3x^2)(dx/dx) +(3y^2)(dy/dx) = 0
(3x^2)(1) + (3y^2)(dy/dx) = 0
3x^2 +(3y^2)y' = 0
y' = -3x^2 / 3y^2 = -x^2 /y^2

----------------------------------------
sinX +cosY = sinXcosY

Differentiate both sides with respect to X,
cosX dX/dX +(-sinY) dY/dX = sinX[-sinY dY/dX] +cosY[cosX dX/dX]
cosX -sinY Y' = -sinXsinY Y' +cosXcosY
-sinY Y' +sinXsinY Y' = cosXcosY -cosX
Y'[sinXsinY -sinY] = cosXcosY -cosX
Y' = [cosXcosY -cosX] / [sinXsinY -sinY]
Y' = [cosX(cosY -1)] / [sinY(sinX -1)]

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