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Someone can tell me what the Partial Derivative of $\frac{d^2z}{dy^2}$ of function $z(x,y)$ if it`s look like this:

$$xy^2+yz^2+xyz+x^2y^2z^2=5$$

I try to solve the first derivative: $$\frac{dz}{dy}=(2yx+z^2z′+xzz′+2x^2yz^2z′)$$

but I am not sure if its okay, Thank you!

BAM
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2 Answers2

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Putting the arguments in the given equality, this is

$$xy^2+yz(x,y)^2+xyz(x,y)+x^2y^2z(x,y)^2=5 .$$

Deriving with respect to $y$ gives

$$2xy + z(x,y)^2 + y 2z(x,y) \frac {\partial z} {\partial y}(x,y) + xz(x,y) + xy \frac {\partial z} {\partial y}(x,y) + x^2 2y z(x,y)^2 + x^2y^2 2 z(x,y) \frac {\partial z} {\partial y}(x,y) = 0 .$$

More nicely written,

$$2xy + z^2 + 2yz \frac {\partial z} {\partial y} + xz + xy \frac {\partial z} {\partial y} + 2x^2 y z^2 + 2x^2 y^2 z \frac {\partial z} {\partial y} = 0 ,$$

whence it follows that

$$\frac {\partial z} {\partial y} = -\frac {2xy + z^2 + xz + 2x^2 yz^2} {2yz + xy + 2x^2 y^2 z} .$$

Of course, one would have to make sure that the denominator exists, but since no domain of definition is given I shall ignore this issue and consider the computation to be purely formal. The expression obtained does not seem to be further simplifiable.

Alex M.
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  • Thank you, but why the derivative of $yz(x,y)^2$ its $z(x,y)^2$ and not $yz(x,y)^2z'$ – BAM May 24 '16 at 10:29
  • @LifeOfPai: You are misreading it: the derivative of $yz^2$ is $1 \cdot z^2 + y \cdot 2z \frac {\partial z} {\partial y}$, because you are dealing with a product so you have to use Leibniz's rule. – Alex M. May 24 '16 at 11:09
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Hint: Note that by the product and the chain rule we have $$ \frac{\partial}{\partial y} y^pz^q = py^{p-1}z^q + y^pqz^{q-1} \frac{\partial}{\partial y} z. $$

Marc
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