The Implicit Differentiation Formulas

We will now look at some formulas for finding partial derivatives of implicit functions. These formulas arise as part of a more complex theorem known as the Implicit Function Theorem which we will get into later.

The Implicit Differentiation Formula for Single Variable Functions

Theorem 1 below will provide us with a method to compute many derivatives of a single variable real-valued functions without having to apply the standard implicit differentiation techniques.

Theorem 1: Suppose that $y = f(x)$ can be rewritten implicitly in the form $F(x, y) = 0$ for all $x \in D(f)$, and suppose that $F$ is differentiable. Then $\frac{dy}{dx} = - \frac{F_x (x, y)}{F_y (x,y)}$ provided that $F_y (x, y) \neq 0$.

For example, if $y = 3x^2 + 2x$, then $0 = 3x^2 + 2y - y$ and so define $F(x, y) = 3x^2 + 2y - y$. Another example is the function $x^3y + 2x + 3y = 2x^2y$, where $0 = x^3y + 2x + 3y - 2x^2y$, and so define $F(x, y) = x^3y + 2x + 3y - 2x^2y$. According to theorem 1, we can find $\frac{dy}{dx}$ if we compute the partial derivative of $F$ with respect to $x$, divide it by the partial derivative of $F$ with respect to $y$ (provided it is not zero), and then negate our result.

Let's look at the proof.

• Proof: Consider a single variable function that can be written implicitly as $F(x, y) = 0$ for all $x \in D(f)$ and suppose that $F $}] is differentiable. We want to find [[$ \frac{dy}{dx}$. We will differentiate both sides of the equation $F(x, y) = 0$ using The Chain Rule Type 1 for Functions of Several Variables. Provided that $\frac{\partial F}{\partial y} \neq 0$ we have that:

(1)

\begin{align} \frac{\partial F}{\partial x} \frac{dx}{dx} + \frac{\partial F}{\partial y} \frac{dy}{dx} = 0 \\ \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} \frac{dy}{dx} = 0 \\ \frac{dy}{dx} = - \frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}} = -\frac{F_x(x, y)}{F_y(x,y)} \quad \blacksquare \end{align}

Now let's look at some examples.

Example 1

Let $3x^2 \cos (xy^3) = 2xye^y$. Find $\frac{dy}{dx}$ using the formula in Theorem 1.

Let $F(x,y) = 3x^2 \cos (xy^3) - 2xye^y$. To apply the formula in Theorem 1, we will need to compute the partial derivatives of $F$ as follows using the product rule:

(2)

\begin{align} \frac{\partial F}{\partial x} = [6x \cos (xy^3) -3x^2y^3 \sin (xy^3)] - 2ye^y \end{align}

(3)

\begin{align} \frac{\partial F}{\partial y} = -9x^3y^2 \sin (xy^3) - 2x[e^y + ye^y] \end{align}

Therefore, using the formula in Theorem 1, we get that:

(4)

\begin{align} \frac{dy}{dx} = - \frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}} = -\frac{[6x \cos (xy^3) -3x^2y^3 \sin (xy^3)] - 2ye^y}{-9x^3y^2 \sin (xy^3) - 2x[e^y + ye^y]} \end{align}

It is suggested that the reader also try to derive the same answer without partial derivatives. The process will be a lot more tedious.

The Implicit Differentiation Formula for Two Variable Functions

Theorem 2: Suppose that $z = f(x, y)$ can be rewritten implicitly in the form $F(x, y, z) = 0$ for all $(x, y) \in D(f)$, and suppose that $F$ is differentiable. Then $\frac{\partial z}{\partial x} = -\frac{F_x (x, y, z)}{F_z (x, y, z)}$ and $\frac{\partial z}{\partial y} = -\frac{F_y (x, y, z)}{F_z (x, y, z)}$ provided that none of the denominators are equal to zero.

• Proof: We will show that $\frac{\partial z}{\partial x} = -\frac{F_x (x, y, z)}{F_z (x, y, z)}$. The other formula is left for the reader to derive.

• Let $z = f(x, y)$ be a function that can be written as $F(x, y, z) = 0$ for all $(x, y) \in D(f)$, and suppose that $F$ is differentiable. Applying the chain rule to both sides of the equation $F(x, y, z) = 0$ we get that:

(5)

\begin{align} \quad \frac{\partial F}{\partial x} \frac{\partial x}{\partial x} + \frac{\partial F}{\partial y} \frac{\partial y}{\partial x} + \frac{\partial F}{\partial z} \frac{\partial z}{\partial x} = 0 \end{align}

• We note that $\frac{\partial x}{\partial x} = \frac{\partial}{\partial x} (x) = 1$ and $\frac{\partial y}{\partial x} = \frac{\partial}{\partial x} (y) = 0$, and so:

(6)

\begin{align} \quad \frac{\partial F}{\partial x} + \frac{\partial F}{\partial z} \frac{\partial z}{\partial x} = 0 \\ \quad \frac{dz}{dx} = - \frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z}} = -\frac{F_x (x, y, z)}{F_z(x, y, z)} \quad \blacksquare \end{align}

Example 2

Let $2x^3yz^2e^z = y e^z\cos z$. Find $\frac{dz}{dx}$.

Let $F(x, y, z) = 2x^3yz^2e^z - y e^z\cos z$. To compute $\frac{\partial z}{\partial x}$ we will need to find $\frac{\partial F}{\partial x}$ and $\frac{\partial F}{\partial z}$ as follows:

(7)

\begin{align} \frac{\partial F}{\partial x} = 6x^2yz^2e^z \end{align}

(8)

\begin{align} \quad \frac{\partial F}{\partial z} = 2x^3y(2ze^z + z^2e^z) - y(e^z \cos z - e^z \sin z) \end{align}

Using the formula in theorem 2, we get that:

(9)

\begin{align} \quad \quad \frac{\partial z}{\partial x} = - \frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z}} = -\frac{6x^2yz^2e^z}{2x^3y(2ze^z + z^2e^z) - y(e^z \cos z - e^z \sin z)} \end{align}