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Problem:

Use implicit differentiation to compute $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ of the function $ x^3 + y^3 +z^3 - 3xyz = 0 $.

What I Got:

$3x^2 + 0 + 3z^2 \cdot \frac{\partial z}{\partial x} - 3y \cdot \frac{\partial z}{\partial x}$

$\frac{\partial z}{\partial x} (3z^2 - 3y) = -3x^2$

$\frac{\partial z}{\partial x} = \frac{-3x^2}{3z^2-3y}$

and

$0 + 3y^2 + 3z^2 \cdot \frac{\partial z}{\partial y} - 3x \cdot \frac{\partial z}{\partial y}$

$\frac{\partial z}{\partial y} (3z^2 - 3x) = -3y^2$

$\frac{\partial z}{\partial y} = \frac{-3y^2}{3z^2-3y}$

For what I have got, is this the correct way to do this question?

Thanks in advance.

Community
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2 Answers2

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You have made an error in implicit partial differentiation of the $−3xyz$ term. Note that$$\frac{\partial(-3xyz)}{\partial x}=-3y\frac{\partial(xz)}{\partial x}$$since $y$ is treated as a constant when differentiating with respect to $x$. Recall that $z$ is actually a function of both $y$ and $x$, i.e. $z=z(x,y)$, so $z$ can't be given the same treatment as $y$.

You can evaluate this using the product rule of differentiation, i.e. $(uv)_x=u_xv+uv_x$, to get$$-3y\frac{\partial(xz)}{\partial x}=-3y\left[z+x\frac{\partial z}{\partial x}\right]$$Similarly when you are differentiating with respect to $y$.

Shubham Johri
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  • Can the answer be achieved with the chain rule? –  Mar 25 '19 at 14:56
  • @user430574 The term $xz$ is a product of two functions of $x$, i.e. $x$ and $z$. You can't differentiate it partially with respect to $x$ without the product rule. Note that if you are aware of the product rule for normal differentiation ($(uv)'=u'v+uv'$), the product rule for partial differentiation can be seen to follow similarly. – Shubham Johri Mar 25 '19 at 15:07
  • then for $\frac {\partial z}{\partial x}$ would it be simplified to: $\frac {\partial z}{\partial x} = \frac{-3x^2}{(3z^2 - 3yz - 3yx)}$ ?? –  Mar 25 '19 at 16:05
  • @user430574 Simplify this: $3x^2 + 0 + 3z^2 \cdot \frac{\partial z}{\partial x} - 3y\left[z+x\frac{\partial z}{\partial x}\right]=0$ to get$$z_x=\dfrac{yz-x^2}{z^2-xy}$$ – Shubham Johri Mar 25 '19 at 16:11
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The easiest way to do this is to simply differentiate the full equation at the beginning, and then create partials from it. So, if you start with $$ x^3 + y^3 + z^3 - 3xyz = 0$$ Then with differentiation, you will get: $$ 3x^2 dx + 3y^2 dy + 3z^2 dz - 3xy dz - 3zx dy - 3 zy dx = 0 $$ Everything divides by 3, giving: $$ x^2 dx + y^2 dy + z^2 dz - xy dz - zx dy - zy dx = 0 $$

You can use this as the starting point for all your partial derivatives.

Now, for each partial differential, just set the other differentials to zero and solve. So, for $\frac{\partial z}{\partial x}$ all instances of $dy$ become $0$ and this becomes: $$ x^2 \partial x + y^2 \cdot 0 + z^2 \partial z - xy \,\partial z - zx \cdot 0 - zy \,\partial x = 0 $$ $$ x^2 \partial x +z^2 \partial z - xy \,\partial z - zy \,\partial x = 0 $$ Now just solve for $\frac{\partial z}{\partial x}$ algebraically. $$ z^2 \partial z - xy \,\partial z = zy \,\partial x - x^2 \partial x $$ $$ \frac{\partial z}{\partial x} = \frac{zy - x^2}{z^2 - xy} $$

You can do the same to get $\frac{\partial z}{\partial y}$.

johnnyb
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