Implicit Partial Differentiation

# Implicit Partial Differentiation

Sometimes a function of several variables cannot neatly be written with one of the variables isolated. For example, consider the following function $x^2y^3z + \cos y \cos z = x^2 \cos x \sin y$. It would be practically impossibly to isolate $z$ let alone any other variable. Fortunately, the concept of implicit differentiation for derivatives of single variable functions can be passed down to partial differentiation of functions of several variables.

Suppose that we wanted to find $\frac{\partial z}{\partial x}$. Then we would take the partial derivatives with respect to $x$ of both sides of this equation and isolate for $\frac{\partial z}{\partial x}$ while treating $y$ as a constant. Applying the product and chain rule where appropriate, we have that:

(1)
\begin{align} \quad \frac{\partial}{\partial x} \left ( x^2y^3z + \cos y \cos z \right) = \frac{\partial}{\partial x} \left ( x^2 \cos x \sin y \right) \\ \quad \quad y^3 \left ( 2xz + x^2 \frac{\partial z}{\partial x} \right ) - \cos y \sin z \frac{\partial z}{\partial x} = (2x \cos x - x^2 \sin x)\sin y \\ \quad \quad 2 xy^3z+ x^2y^3 \frac{\partial z}{\partial x} - \cos y \sin z \frac{\partial z}{\partial x} = (2x \cos x - x^2 \sin x)\sin y \\ x^2y^3 \frac{\partial z}{\partial x} - \cos y \sin z \frac{\partial z}{\partial x} = (2x \cos x - x^2 \sin x)\sin y - 2 xy^3z \\ \frac{\partial z}{\partial x} \left (x^2y^3- \cos y \sin z \right ) = (2x \cos x - x^2 \sin x)\sin y - 2 xy^3z \\ \frac{\partial z}{\partial x} = \frac{(2x \cos x - x^2 \sin x)\sin y - 2 xy^3z}{x^2y^3- \cos y \sin z} \end{align}

Therefore we have implicitly solved for $\frac{\partial z}{\partial x}$. We will now look at some more examples.

## Example 1

Let $z^2 = x^2 + y^2$. Find $\frac{\partial z}{\partial y}$.

Let's take the partial derivatives of both sides with respect to $y$ and treat $x$ as a constant.

(2)
\begin{align} \frac{\partial}{\partial y} z^2 = \frac{\partial}{\partial y} \left ( x^2 + y^2 \right )\\ 2z \frac{\partial z}{\partial y} = 2y \\ \frac{\partial z}{\partial y} = \frac{y}{z} \end{align}

## Example 2

Let $z \cos z = x^2 y^3 + z$. Find $\frac{\partial z}{\partial x}$.

Let's take the partial derivatives of both sides with respect to $x$ and treat $y$ as a constant.

(3)
\begin{align} \quad \frac{\partial}{\partial x} \left ( z \cos z \right )= \frac{\partial}{\partial x} \left ( x^2 y^3 + z \right ) \\ \quad \frac{\partial z }{\partial x} \cos z -z \sin z \frac{\partial z}{\partial x} = 2x y^3 + \frac{\partial z}{\partial x} \\ \quad \frac{\partial z }{\partial x} \cos z -z \sin z \frac{\partial z}{\partial x} - \frac{\partial z}{\partial x} = 2xy^3 \\ \quad \frac{\partial z}{\partial x} \left (\cos z - z\sin z - 1\right ) = 2xy^3 \\ \quad \frac{\partial z}{\partial x} = \frac{2xy^3}{\cos z - z \sin z - 1} \end{align}