Implicit differentiation for functions of several variables

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Multivariable Functions & Partial Derivatives

A function of two variables $f: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ assigns a unique real number to each ordered pair in its domain, e.g., $Area = \frac{1}{2}bh$, $z = 2x + 3y$, and $f(x, y) = x^2 + y^2$.

Partial derivatives are derivatives of a function of multiple variables, say $f(x_1, x_2, \cdots, x_n)$, with respect to one of those variables, with the other variables held constant. They measure how the function changes as one variable changes, while keeping the others constant. $\frac{\partial f}{\partial x}(x_0, y_0) = \lim_{\Delta x \to 0} \frac{f(x_0+\Delta x, y_0)-f(x_0, y_0)}{\Delta x}$

Geometrically, the partial derivative $\frac{\partial f}{\partial x}(x_0, y_0)$ at a point $(x_0, y_0)$ can be interpreted as the slope of the tangent line to the curve formed by the intersection of the surface $z = f(x,y)$ and the plane $y = y_0$. Similarly, $\frac{\partial f}{\partial y}(x_0, y_0)$ represents the slope of the tangent line to the curve formed by the intersection of the surface defined by $z = f(x,y)$ and the plane $x = x_0$.

Directional Derivatives & The Gradient

The directional derivative $\frac{dw}{ds}\bigg\vert{}_{\vec{u}}$ is the rate of change of the function $w$ as we move in the direction of a unit vector $\vec{u}$:

$\frac{dw}{ds}\bigg\vert{}_{\vec{u}} = \nabla w \cdot \vec{u} = \vert{}\nabla w\vert{} \cdot \vert{}\vec{u}\vert{}\cos(\theta) = \vert{}\nabla w\vert{}\cos(\theta)$ where $\theta$ is the angle between the gradient and the given unit vector $\vec{u}$.

• Maximum (Steepest Ascent): The directional derivative is maximized when $\cos(\theta) = 1 \iff \theta = 0$. This means $\vec{u}$ is aligned exactly with the gradient $\nabla w$. The gradient points in the direction of the steepest ascent (fastest increase) at a given point.
• Minimum (Steepest Descent): The directional derivative is minimized when $\cos(\theta) = -1 \iff \theta = 180^\circ$. This means $\vec{u}$ points in the exact opposite direction of the gradient.
• Zero Change: $\frac{dw}{ds}\bigg\vert{}_{\vec{u}} = 0 \iff \cos(\theta) = 0 \iff \theta = 90^\circ$. Moving orthogonally ($\vec{u} \perp \nabla w$) to the gradient results in no change in the function’s value.

The gradient vector is defined as $\nabla w = \langle w_x, w_y, w_z \rangle = \langle \frac{\partial w}{\partial x}, \frac{\partial w}{\partial y}, \frac{\partial w}{\partial z} \rangle$. Because there is zero change in the function’s value along a level curve or level surface, the rate of change is zero, meaning the gradient is always perpendicular (orthogonal) to the level curves.

The Chain Rule

Let $w = f(x, y, z)$ where $x = x(t)$, $y = y(t)$, and $z = z(t)$ are parametrised functions of another variable $t$. The derivative of $w$ with respect to $t$ is calculated via the Chain Rule: $\frac{dw}{dt} = \frac{\partial w}{\partial x}\frac{dx}{dt} + \frac{\partial w}{\partial y}\frac{dy}{dt} + \frac{\partial w}{\partial z}\frac{dz}{dt}$

This can be neatly rewritten using the dot product of the gradient and the velocity vector: $\frac{dw}{dt} = \nabla w \cdot \frac{d\vec{r}}{dt}$ where $\frac{d\vec{r}}{dt} = \langle \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \rangle$.

Implicit Differentiation: An Overview

In mathematics, some equations involving two variables, x and y, do not explicitly define y as a function of x. This means that even though y can be theoretically expressed as a function of x implicitly, the equation might be too complex or even impossible to rearrange in a way that solves for y explicitly in terms of x. These equations are said to be defined implicitly.

For example, consider an equation like $x^2 + y^2 = 1$. This equation implicitly defines a relationship between x and y, but it does not express y directly as a function of x (such as $y = f(x)$). However, there might still exist a function $y = f(x)$ that satisfies this equation for each x.

The technique of implicit differentiation allows us to find the derivative of y with respect to x (denoted as $\frac{dy}{dx}$) without needing to solve the equation explicitly for y.

Examples

• Implicit Differentiation of a Circle. Consider the equation of a circle with a radius of 1 centered at the origin: $x^2 + y^2 = 1$.

To find the derivative $\frac{dy}{dx}$ we can differentiate both sides of the equation with respect to x, applying the rules of differentiation: $\frac{d}{dx}(x^2+y^2=1) \implies[\text{For the term y², we use the chain rule}] 2x+2y\frac{dy}{dx}=0$.
Isolate $\frac{dy}{dx}$: $\frac{dy}{dx}=\frac{-x}{y}.$
It gives the slope of the tangent line to the circle at any point $(x, y)$ on the circle.

• Differentiating the Inverse Sine Function. Consider the function $y = \sin^{-1}(x)$.

1. To differentiate this implicitly, we start by rewriting the equation in a form that does not involve the inverse function: $\sin(y) = x.$
2. Now, differentiate both sides with respect to x: $\frac{d}{dx}\sin(y) = 1 \implies \cos(y)\frac{dy}{dx}=1 \implies \frac{dy}{dx} = \frac{1}{cos(y)}$.
3. Using the Pythagorean identity $\sin^2(y) + \cos^2(y) = 1$, we know $\cos(y) = \sqrt{1 - \sin^2(y)}$. Since $\sin(y) = x$, this becomes $\sqrt{1 - x^2}$.
4. Therefore, $\frac{d}{dx}(\sin^{-1}(x)) = \frac{1}{\sqrt{1 - x^2}}$.

Calculating Partial Derivatives for Implicit Functions

We can apply this logic to 3D surfaces. Let’s find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ for the implicit surface: $x^2 + y^4 - z^3 + 3xy^2 = 8$

1. Rewrite the Implicit Equation as a single function $f(x,y,z)$. First, let’s rewrite the equation in a form that’s easier to differentiate: $f(x, y, z) = x^2 + y^4 - z^3 + 3xy^2 - 8 = 0$. It defines z implicitly as a function of x and y.

2. Compute the Partial derivatives of $f$
   Next, we need to calculate the partial derivatives of $f(x, y, z)$ with respect to x, y, and z respectively:
   $f_x = \frac{\partial f}{\partial x} = 2x + 3y^2$, $f_y = \frac{\partial f}{\partial y} = 4y^3 + 6xy$, $f_z = \frac{\partial f}{\partial z} = -3z^2$

3. Implicit Differentiation. To compute $\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}$, we use implicit differentiation. The key idea is to differentiate the implicit equation $f(x, y, z) = 0$ with respect to the desired variable while treating z as a function of x and y, $z = z(x, y)$.

4. Partial derivative $\frac{\partial z}{\partial x}$. Differentiate f(x, y, z) = 0 with respect to x (treating $y$ as a constant, so $\frac{\partial y}{\partial x} = 0$): $\frac{\partial f}{\partial x}\frac{\partial x}{\partial x} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial x}+ \frac{\partial f}{\partial z}\frac{\partial z}{\partial x}$
   Since y is fixed, we have: $\frac{\partial x}{\partial x}=1, \frac{\partial y}{\partial x} = 0$.
   So, the previous equation simplifies to: $\frac{\partial f}{\partial x}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial x} = f_x+f_z\frac{\partial z}{\partial x}=0⇒[\text{Solve for} \frac{\partial z}{\partial x}] \frac{\partial z}{\partial x}=z_x = \frac{-f_x}{f_z}$
   $\frac{\partial z}{\partial x}= z_x = \frac{-f_x}{f_z} =[\text{Plug in the values we found in step 2}] \frac{-(2x+3y^2)}{-3z^2} = \frac{2x+3y^2}{3z^2}$.

5. Partial derivative $\frac{\partial z}{\partial y}$. Similarly, to find $\frac{\partial z}{\partial y}$, we differentiate f(x, y, z) = 0 with respect to y: $\frac{\partial f}{\partial x}\frac{\partial x}{\partial y} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial y}+ \frac{\partial f}{\partial z}\frac{\partial z}{\partial y}$
   Since x is fixed, we have: $\frac{\partial x}{\partial y}=0, \frac{\partial y}{\partial y} = 1$
   So, the equation simplifies to: $f_x·0 + f_y·1 + f_z·\frac{\partial z}{\partial y} = 0↭ f_y + f_z·\frac{\partial z}{\partial y} = 0$. Solve for $\frac{\partial z}{\partial y}$: $\frac{\partial z}{\partial y} = \frac{-f_y}{f_z}$
   $\frac{\partial z}{\partial y}= z_y = \frac{-f_y}{f_z} =[\text{Plug in the values we found in step 2}] \frac{-( 4y^3+6xy)}{-3z^2} = \frac{4y^3+6xy}{3z^2}$.

• Given the implicit function: $x^2\sin(2y - 5z) = 1 + y\cos(6zx)$. We want to find the partial derivatives of z with respect to x and y, denoted as $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$

1. Rewrite the Implicit Equation. Bring all terms to one side to define a continuous multivariable function $f(x, y, z) = 0$, namely $f(x, y, z) = x^2\sin(2y - 5z) - 1 - y\cos(6zx) = 0$. This defines $z$ implicitly as a function of the independent variables $x$ and $y$, such that $z = z(x, y)$.

2. Compute the Partial derivatives of $f(x, y, z)$. Next, we need to calculate the partial derivatives of $f(x, y, z)$ with respect to x, y, and z respectively:
   $\star$ For $f_x$: Apply the product rule to the first term and the chain rule to the third term.
   $f_x = \frac{\partial}{\partial x}[x^2\sin(2y - 5z)] - \frac{\partial}{\partial x}[y\cos(6zx)]$
   $f_x = 2x\sin(2y - 5z) - y[-\sin(6zx) \cdot (6z)] = 2x\sin(2y - 5z) + 6yz\sin(6zx)$
   $\star$ For $f_y$: $f_y = \frac{\partial}{\partial y}[x^2\sin(2y - 5z)] - \frac{\partial}{\partial y}[y\cos(6zx)]$
   $f_y = x^2[\cos(2y - 5z) \cdot (2)] - (1)\cos(6zx) = 2x^2\cos(2y - 5z) - \cos(6zx)$
   $\star$ For $f_z$: $f_z = \frac{\partial}{\partial z}[x^2\sin(2y - 5z)] - \frac{\partial}{\partial z}[y\cos(6zx)]$
   $f_z = x^2[\cos(2y - 5z) \cdot (-5)] - y[-\sin(6zx) \cdot (6x)] = -5x^2\cos(2y - 5z) + 6xy\sin(6zx)$.

3. Implicit Differentiation.
   To compute $\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}$, we use implicit differentiation. The key idea is to differentiate the implicit equation $f(x, y, z) = 0$ with respect to the desired variable while treating z as a function of x and y, $z = z(x, y)$.

4. Partial derivative $\frac{\partial z}{\partial x}$
   Differentiate $f(x, y, z) = 0$ with respect to x: $\frac{\partial f}{\partial x}\frac{\partial x}{\partial x} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial x}+ \frac{\partial f}{\partial z}\frac{\partial z}{\partial x} = 0$
   Since y is fixed, we have: $\frac{\partial x}{\partial x}=1, \frac{\partial y}{\partial x} = 0$.
   So, the previous equation simplifies to: $\frac{\partial f}{\partial x}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial x} = f_x+f_z\frac{\partial z}{\partial x}=0⇒[\text{Solve for} \frac{\partial z}{\partial x}] \frac{\partial z}{\partial x}=z_x = \frac{-f_x}{f_z}$
   $\frac{\partial z}{\partial x}= z_x = \frac{-f_x}{f_z} =[\text{Plug in the values}] \frac{-(2x\sin(2y -5z) +6yz\sin(6zx))}{-5x^2\cos(2y -5z) +6yxsin(6zx)} = \frac{2xsin(2y -5z) +6yzsin(6zx)}{5x^2cos(2y -5z) -6yxsin(6zx)}$.

5. Partial derivative $\frac{\partial z}{\partial y}$. Similarly, to find $\frac{\partial z}{\partial y}$, we differentiate f(x, y, z) = 0 with respect to y: $\frac{\partial f}{\partial x}\frac{\partial x}{\partial y} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial y}+ \frac{\partial f}{\partial z}\frac{\partial z}{\partial y}$
   Since x is fixed, we have: $\frac{\partial x}{\partial y}=0, \frac{\partial y}{\partial y} = 1$
   So, the equation simplifies to: $f_x·0 + f_y·1 + f_z·\frac{\partial z}{\partial y} = 0↭ f_y + f_z·\frac{\partial z}{\partial y} = 0$. Solve for $\frac{\partial z}{\partial y}$: $\frac{\partial z}{\partial y} = \frac{-f_y}{f_z}$
   $\frac{\partial z}{\partial y}= z_y = \frac{-f_y}{f_z} =[\text{Plug in the values}] \frac{-(2x^2cos(2y -5z) -cos(6zx))}{-5x^2cos(2y -5z) +6yxsin(6zx)} = \frac{2x^2cos(2y -5z)-cos(6zx)}{5x^2cos(2y -5z) -6yxsin(6zx)}$·

• Given the implicit function: $x^4 + y^4 + z^4 + x^2y^2z^2 = 0$. We want to find the partial derivatives of y with respect to z, denoted as $\frac{\partial y}{\partial z}$.

1. Define the Function and Relationships.
The equation is already set to zero, so $f(x, y, z) = x^4 + y^4 + z^4 + x^2y^2z^2 = 0$.
Because we are looking for $\frac{\partial y}{\partial z}$, we are treating $y$ as the dependent variable driven by the independent variables $x$ and $z$. This means $y = y(x, z)$.

2. Compute the Necessary Partial Derivatives.
To find $\frac{\partial y}{\partial z}$, we only strictly need $f_y$ and $f_z$ (though calculating $f_x$ is good practice for verifying the full gradient).
$f_x = \frac{\partial f}{\partial x} = 4x^3 + 2xy^2z^2$, $f_y = \frac{\partial f}{\partial y} = 4y^3 + 2x^2yz^2$, $f_z = \frac{\partial f}{\partial z} = 4z^3 + 2x^2y^2z$

3. Assemble the Implicit Derivative.
Differentiate f(x, y, z) = 0 with respect to z: $\frac{\partial f}{\partial x}\frac{\partial x}{\partial z} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial z}+ \frac{\partial f}{\partial z}\frac{\partial z}{\partial z}= 0$
Since x is fixed, we have: $\frac{\partial x}{\partial z}=0, \frac{\partial z}{\partial z} = 1$.
The chain rule dictates: $\frac{\partial f}{\partial y}\frac{\partial y}{\partial z} + \frac{\partial f}{\partial z}(1) = 0 \implies \frac{\partial y}{\partial z} = -\frac{f_z}{f_y}$
Plug in the values and factor out common terms to simplify: $\frac{\partial y}{\partial z} = -\frac{4z^3 + 2x^2y^2z}{4y^3 + 2x^2yz^2}$
$\frac{\partial y}{\partial z} = -\frac{2z(2z^2 + x^2y^2)}{2y(2y^2 + x^2z^2)} = -\frac{z(2z^2 + x^2y^2)}{y(2y^2 + x^2z^2)}$

4. A Mathematical Nuance: 🎯 While the algebraic mechanics of implicit differentiation work perfectly here, the real-number domain of the function $x^4 + y^4 + z^4 + x^2y^2z^2 = 0$ is exactly one point: $(0,0,0)$, because all terms are strictly non-negative.
A sum of non‑negative real numbers equals zero only if each term is individually zero.
At this specific coordinate, evaluating $\frac{\partial y}{\partial z}$ results in division by zero ($\frac{0}{0}$), meaning the derivative is undefined everywhere in the real geometric space.

The formula $\frac{\partial y}{\partial z} =- \frac{f_z}{f_y}$ comes from the assumption that the level set $f(x, y, z) = 0$ is a smooth 2‑dimensional surface near the point of interest. For that to be true, the gradient must be nonzero: $\nabla f(x, y, z) \ne 0$. However, at $(0, 0, 0), f_x = f_y = f_z = 0$. So the gradient vanishes. This means the level set is not a smooth surface there — it is a singular point.
Geometrically, there is no curve to differentiate along. There is no slope.

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