Implicit Differentiation of Partial Derivatives

**Belowzero78** · April 7th 2010, 06:17 PM

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

**Chris L T521** · April 7th 2010, 07:18 PM

Originally Posted by Belowzero78

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

How much have you tried?

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

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

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

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

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

Can you take it from here?

**Belowzero78** · April 7th 2010, 07:46 PM

Originally Posted by Chris L T521

How much have you tried?

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

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

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

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

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

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?

**HallsofIvy** · April 8th 2010, 04:31 AM

In the last two terms? There is no y in 216xz.

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

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

**Belowzero78** · April 13th 2010, 01:51 PM

Here is what i got. Can anyone please confirm?

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

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 $y = \frac{3}{2}$ . I's unsure of why the answer is -1/2. Im missing a negative sign somewhere.... since im getting 1/2.

**Belowzero78** · April 13th 2010, 02:07 PM

SOLVED. I figured out my mistakes... thanks everybody.

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