# How do you differentiate x^3 yz^2+w^2 x^2 y-x^3+wsiny=44?

Aug 26, 2015

$\frac{\partial w}{\partial x} = \frac{3 {x}^{2} \left(1 - y {z}^{2}\right) - 2 x y {w}^{2}}{2 {x}^{2} y w + \sin y}$
$\frac{\partial w}{\partial y} = - \frac{{x}^{2} \left(x {z}^{2} + {w}^{2}\right) + w \cos y}{2 {x}^{2} + \sin y}$
$\frac{\partial w}{\partial z} = - \frac{2 {x}^{3} y z}{2 {x}^{2} y w + \sin y}$

#### Explanation:

I'll assume that $w$ is your dependent variable and that $x$, $y$, and $z$ are all independent variables.

You can calculate the partial derivative of $w$ with respect to each independent variable by using implicit differentiation. Keep in mind, when you differentiate with respect to $x$, $y$ and $z$ are treated like constants.

When you differentiate with respect to $y$, $x$ and $z$ are treated like constants. And so on.

On the other hand, $w$ is treated like a function that depends on all three independent variables. This means that you're going to use the chain rule whenever you derive $w$ with respect to an independent variable.

So, let's start with the partial derivative of $w$ with respect to $x$. You're dealing with an equation, which means that you need to differentiate both sides with respect to $x$.

$\frac{\partial}{\partial x} \left({x}^{3} y {z}^{2} + {w}^{2} {x}^{2} y - {x}^{3} + w \sin y\right) = \frac{d}{\mathrm{dx}} \left(44\right)$

Now, you can calculate $\frac{\partial}{\partial x} \left(y {x}^{2} {w}^{2}\right)$ by using the product rule. Remember that $y$ is treated like a constant

$\frac{\partial}{\partial x} \left(y {x}^{2} {w}^{2}\right) = y \cdot \left[\frac{d}{\mathrm{dx}} \left({x}^{2}\right) \cdot {w}^{2} + {x}^{2} \cdot \frac{\partial}{\partial x} \left({w}^{2}\right)\right]$

$\frac{\partial}{\partial x} \left(y {x}^{2} {w}^{2}\right) = y \cdot \left(2 x \cdot {w}^{2} + {x}^{2} \cdot 2 w \cdot \frac{\partial w}{\partial x}\right)$

$\frac{\partial}{\partial x} \left(y {x}^{2} {w}^{2}\right) = 2 x y {w}^{2} + 2 {x}^{2} y w \frac{\partial w}{\partial x}$

Plug this back into the calculation with respect to $x$ to get

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

Rearrange this to isolate $\frac{\partial w}{\partial x}$ on one side

$\frac{\partial w}{\partial x} \left(2 {x}^{2} y w + \sin y\right) = + 3 {x}^{2} - 2 x y {w}^{2} - 3 {x}^{2} y {z}^{2}$

$\frac{\partial w}{\partial x} = \textcolor{g r e e n}{\frac{3 {x}^{2} \left(1 - y {z}^{2}\right) - 2 x y {w}^{2}}{2 {x}^{2} y w + \sin y}}$

Next, take the partial derivative of $w$ with respect to $y$.

$\frac{\partial}{\partial y} \left({x}^{3} y {z}^{2} + {w}^{2} {x}^{2} y - {x}^{3} + w \sin y\right) = \frac{d}{\mathrm{dy}} \left(44\right)$

This will get you

${x}^{3} {z}^{2} + {x}^{2} \cdot \left[\frac{\partial}{\partial y} \left({w}^{2}\right) \cdot y + {w}^{2} \cdot \frac{d}{\mathrm{dy}} \left(y\right)\right] - 0 + \left[\frac{\partial}{\partial y} \left(w\right) \cdot \sin y + w \cdot \frac{d}{\mathrm{dy}} \left(\sin y\right)\right] = 0$

${x}^{3} {z}^{2} + {x}^{2} \cdot \left(2 w \cdot \frac{\partial w}{\partial y} + {w}^{2}\right) + \frac{\partial w}{\partial y} \cdot \sin y + w \cdot \cos y = 0$

${x}^{3} {z}^{2} + 2 {x}^{2} w \frac{\partial w}{\partial y} + {x}^{2} {w}^{2} + \frac{\partial w}{\partial y} \cdot \sin y + w \cos y = 0$

Once again, rearrange to get $\frac{\partial w}{\partial y}$ isoltated on one side

$\frac{\partial w}{\partial y} \left(2 {x}^{2} w + \sin y\right) = - {x}^{3} {z}^{2} - {x}^{2} {w}^{2} - w \cos y$

$\frac{\partial w}{\partial y} = \textcolor{g r e e n}{- \frac{{x}^{2} \left(x {z}^{2} + {w}^{2}\right) + w \cos y}{2 {x}^{2} + \sin y}}$

Finally, take the partial derivative of $w$ with respect to $z$. This one will be easier to calculate because you won't use the product rule

$\frac{\partial}{\partial z} \left({x}^{3} y {z}^{2} + {w}^{2} {x}^{2} y - {x}^{3} + w \sin y\right) = \frac{d}{\mathrm{dz}} \left(44\right)$

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

Rearrange to get $\frac{\partial w}{\partial z}$ alone on one side

$\frac{\partial w}{\partial z} \left(2 {x}^{2} y w + \sin y\right) = - 2 {x}^{3} y z$

$\frac{\partial w}{\partial z} = \textcolor{g r e e n}{- \frac{2 {x}^{3} y z}{2 {x}^{2} y w + \sin y}}$