Implicit Partial Deriviative

**Gotovina7** · April 19th 2008, 02:47 PM

Stuck on this one also

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

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

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

Thanks!

**mr fantastic** · April 19th 2008, 03:07 PM

Originally Posted by Gotovina7

Stuck on this one also

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

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

compute $\frac{dy}{dz}$ 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.

**Gotovina7** · April 19th 2008, 03:08 PM

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?

**mr fantastic** · April 19th 2008, 03:16 PM

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 ....

**TheEmptySet** · April 19th 2008, 03:19 PM

Originally Posted by Gotovina7

Stuck on this one also

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

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

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

Thanks!

the code for a partial is \partial y

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

eval @ (4,-3,1)

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

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

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

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