1. ## Implicit Partial Deriviative

Stuck on this one also

Given that y=y(x,z) is defined implicitly by the relation

$\displaystyle y^3 + z\sqrt{x} - ln(xz^{2}-y) = \frac{16}{x^2} +24$

compute $\displaystyle \frac{dy}{dz}$ at the point (4,-3,1)

Thanks!

2. Originally Posted by Gotovina7
Stuck on this one also

Given that y=y(x,z) is defined implicitly by the relation

$\displaystyle y^3 + z\sqrt{x} - ln(xz^{2}-y) = \frac{16}{x^2} +24$

compute $\displaystyle \frac{dy}{dz}$ at the point (4,-3,1)

Thanks!
Treat x as a constant and differentiate implicitly as you would in the single variable case. Half the answer: the derivative of the right hand side is zero.

3. Originally Posted by mr fantastic
Treat x as a constant and differentiate implicitly as you would in the single variable case. Half the answer: the derivative of the right hand side is zero.
Ok, should I split up that ln?

4. Originally Posted by Gotovina7
Ok, should I split up that ln?
xz^2 - y cannot be factorised. Since you're not logging a pure product, you can't split the log up ....

5. Originally Posted by Gotovina7
Stuck on this one also

Given that y=y(x,z) is defined implicitly by the relation

$\displaystyle y^3 + z\sqrt{x} - ln(xz^{2}-y) = \frac{16}{x^2} +24$

compute $\displaystyle \frac{dy}{dz}$ at the point (4,-3,1)

Thanks!
the code for a partial is \partial y

$\displaystyle 3y^2\frac{\partial y}{\partial z}+\sqrt{x}-\frac{2xz-\frac{\partial y}{\partial z}}{xz^2-y}=0$

eval @ (4,-3,1)

$\displaystyle 3(-3)^2\frac{\partial y}{\partial z}+\sqrt{4}-\frac{2(4)(1)-\frac{\partial y}{\partial z}}{(4)(1)^2-(-3)}=0$

$\displaystyle 27 \frac{\partial y}{\partial z} +2 -\frac{8-\frac{\partial y}{\partial z}}{7}=0$

$\displaystyle 189 \frac{\partial y}{\partial z} +14 -8 +\frac{\partial y}{\partial z}=0 \iff 190 \frac{\partial y}{\partial z} =-6 \iff \frac{\partial y}{\partial z}=\frac{-3}{95}$