# Implicit Differentiation of Partial Derivatives

• Apr 7th 2010, 06:17 PM
Belowzero78
Implicit Differentiation of Partial Derivatives
Question: $\displaystyle ln(x^5z^{216}) + 9 (x - 1)y + 216xz + 64y^3z = 0$ defines $\displaystyle y$ implicitly as a function of $\displaystyle x$ and $\displaystyle z$; that is $\displaystyle y =y(x,z)$. Compute $\displaystyle \frac{\partial y}{\partial z} at (x,z) = (1,-1)$.
• Apr 7th 2010, 07:18 PM
Chris L T521
Quote:

Originally Posted by Belowzero78
Question: $\displaystyle ln(x^5z^{216}) + 9 (x - 1)y + 216xz + 64y^3z = 0$ defines $\displaystyle y$ implicitly as a function of $\displaystyle x$ and $\displaystyle z$; that is $\displaystyle y =y(x,z)$. Compute $\displaystyle \frac{\partial y}{\partial z} at (x,z) = (1,-1)$.

How much have you tried?

$\displaystyle \frac{\partial}{\partial z}\left[\ln(x^5z^{216})\right]$ should be routine.

$\displaystyle \frac{\partial }{\partial z}\left[9(x-1)y\right]=9(x-1)\frac{\partial y}{\partial z}$

$\displaystyle \frac{\partial}{\partial z}\left[216xz\right]$ should be routine.

$\displaystyle \frac{\partial}{\partial z}\left[64y^3z\right]$ requires product rule.

Combine all these results, and then solve for $\displaystyle \frac{\partial y}{\partial z}$

Can you take it from here?
• Apr 7th 2010, 07:46 PM
Belowzero78
Quote:

Originally Posted by Chris L T521
How much have you tried?

$\displaystyle \frac{\partial}{\partial z}\left[\ln(x^5z^{216})\right]$ should be routine.

$\displaystyle \frac{\partial }{\partial z}\left[9(x-1)y\right]=9(x-1)\frac{\partial y}{\partial z}$

$\displaystyle \frac{\partial}{\partial z}\left[216xz\right]$ should be routine.

$\displaystyle \frac{\partial}{\partial z}\left[64y^3z\right]$ requires product rule.

Combine all these results, and then solve for $\displaystyle \frac{\partial y}{\partial z}$

Can you take it from here?

Could please show me what happens with the y in the last two terms? When i evaluate the derivative wont I have a problem?
• Apr 8th 2010, 04:31 AM
HallsofIvy
In the last two terms? There is no y in 216xz.

$\displaystyle \frac{\partial 64y^3z}{\partial x}= 64(3y^2)\frac{\partial y}{\partial x} z= 192y^2z \frac{\partial y}{\partial x}$.

$\displaystyle \frac{\partial 64y^3z}{\partial z}= $$\displaystyle 64(3y^2)\frac{\partial y}{\partial z} z+ 64y^3=$$\displaystyle 192y^2z\frac{\partial y}{\partial z}+ 64y^3$.
• Apr 13th 2010, 01:51 PM
Belowzero78
Here is what i got. Can anyone please confirm?

$\displaystyle \frac{\partial y}{\partial z} = \frac{\frac{216}{z} - 216x -64y^3}{9x -9 + 192zy^2}$

I am confused onto find what z is. Could someone please explain. I know that y is a function of x and z, but how do i find it?

Thanks!

Nevermind, i found $\displaystyle y = \frac{3}{2}$. I's unsure of why the answer is -1/2. Im missing a negative sign somewhere.... since im getting 1/2.
• Apr 13th 2010, 02:07 PM
Belowzero78
SOLVED. I figured out my mistakes... thanks everybody.