Implicit Derivative Calculator

What is an Implicit Derivative Calculator?

An Implicit Derivative Calculator differentiates equations where \(y\) is defined implicitly by a relation \(F(x,y)=0\) rather than an explicit function \(y=f(x)\). By treating \(y\) as a function of \(x\) during differentiation and applying the chain rule, the calculator isolates \(\dfrac{dy}{dx}\) (and, when requested, higher-order derivatives). This is essential for curves that are hard—or impossible—to solve explicitly for \(y\), such as circles, lemniscates, or algebraic curves with multiple branches. The tool presents each algebraic step, flags vertical tangents where \(\dfrac{dy}{dx}\) is undefined, and can evaluate the derivative at specific points on the curve.

About the Implicit Derivative Calculator

Given a differentiable relation \(F(x,y)=0\), implicit differentiation yields \[ F_x(x,y) + F_y(x,y)\,\frac{dy}{dx} = 0 \quad\Rightarrow\quad \frac{dy}{dx} = -\,\frac{F_x}{F_y}\quad (F_y\ne0). \] The calculator can work directly from \(F\) (via partial derivatives) or expand \(y(x)\) chain-rule terms if the input is given as an equation. It simplifies the expression for \(\dfrac{dy}{dx}\), checks denominators for zeros (possible vertical tangents), and supports second derivatives by differentiating \(\dfrac{dy}{dx}\) implicitly again. When a point \((x_0,y_0)\) is supplied, the tool verifies that it satisfies the relation and then substitutes it to return a numerical slope and, if requested, curvature via \(\dfrac{d^2y}{dx^2}\). Exact arithmetic is kept for rational data; otherwise high-precision decimals are shown.

How to Use this Implicit Derivative Calculator

1. Enter the relation \(F(x,y)=0\) (e.g., \(x^2+y^2-25=0\)).
2. (Optional) Enter a point \((x_0,y_0)\) on the curve to evaluate the slope numerically.
3. Choose derivative order: first derivative \(\dfrac{dy}{dx}\) or also compute \(\dfrac{d^2y}{dx^2}\).
4. Run the calculation to see the partials, the solved formula for \(\dfrac{dy}{dx}\), and any evaluated values at your point.
5. Copy the simplified LaTeX to include in notes, reports, or solutions.

Core Formulas (LaTeX)

From a relation \(F(x,y)=0\): \[ F_x + F_y\,\frac{dy}{dx}=0 \ \Rightarrow\ \frac{dy}{dx}=-\frac{F_x}{F_y},\quad F_y\neq0. \]

Chain rule view: \[ \frac{d}{dx}\,G(x,y(x)) = G_x + G_y\,\frac{dy}{dx}. \]

Second derivative (schematic): \[ \frac{d^2y}{dx^2} = \frac{d}{dx}\!\left(\frac{dy}{dx}\right) = \frac{\partial}{\partial x}\!\left(\frac{dy}{dx}\right) + \frac{\partial}{\partial y}\!\left(\frac{dy}{dx}\right)\frac{dy}{dx}. \]

Examples (Illustrative)

Example 1 — Circle \(x^2+y^2=25\)

Differentiate: \(2x + 2y\,\dfrac{dy}{dx}=0 \Rightarrow \dfrac{dy}{dx}=-\dfrac{x}{y}\). At \((3,4)\): slope \(m=-\tfrac{3}{4}\). Second derivative: \[ y'=-\frac{x}{y}\ \Rightarrow\ y''=-\frac{y - x y'}{y^2}=-\frac{y + \frac{x^2}{y}}{y^2}=-\frac{x^2+y^2}{y^3}=-\frac{25}{y^3}; \] at \((3,4)\): \(y''=-\tfrac{25}{64}\).

Example 2 — Folium \(x^3+y^3=9xy\)

Differentiate: \(3x^2 + 3y^2 y' = 9(y + x y')\). Solve: \[ y'(3y^2 - 9x) = 9y - 3x^2 \ \Rightarrow\ y'=\frac{9y - 3x^2}{3y^2 - 9x}=\frac{3y - x^2}{y^2 - 3x}. \]

Example 3 — Exponential \(e^{xy} + x - y = 0\)

Differentiate: \(e^{xy}(x y' + y) + 1 - y' = 0\). Collect \(y'\): \[ y'\,(x e^{xy} - 1) = -\,\big(y e^{xy} + 1\big) \ \Rightarrow\ y'=\frac{1 + y e^{xy}}{1 - x e^{xy}}. \]

FAQs

What is implicit differentiation?

A method to differentiate relations \(F(x,y)=0\) by treating \(y\) as \(y(x)\) and applying the chain rule to isolate \(\dfrac{dy}{dx}\).

When is \(\dfrac{dy}{dx}\) undefined?

Where \(F_y=0\) while \(F_x\ne0\); these are vertical tangents or points where slope is not finite.

Can the calculator evaluate at a point?

Yes—after verifying the point lies on the curve, it substitutes to give numerical slopes (and second derivatives if requested).

How do I get the second derivative?

The tool differentiates the first-derivative expression implicitly again and simplifies to return \(\dfrac{d^2y}{dx^2}\).

What if the relation has multiple branches?

The derivative formula applies on any branch satisfying \(F_y\ne0\); sign/values depend on the chosen point.

Does this work in 3D?

For surfaces \(F(x,y,z)=0\), you can compute partials (e.g., \(\dfrac{\partial z}{\partial x}=-F_x/F_z\)) similarly, provided the relevant partial is nonzero.