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  • Executing Implicit Differentiation

  • Deducing the Tangent Equation of a Curve via Implicit Differentiation

  • Formulating a Curve's Normal Equation Through Implicit Differentiation

  • Essential Insights on Implicit Differentiation

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Understanding Implicit differentiation: Definition, Examples & Formula

Implicitly defined functions involve two variables, x and y, where y is not separated on either side of the equation. In some cases, equations are too complex to simply rearrange for solving y as a function of x, which complicates standard differentiation. Implicit differentiation offers a technique to differentiate y in relation to x (dy/dx) without having to solve for y first.

This technique is also applicable in determining the gradient and curvature of curves depicted by parametric equations, typical of implicit function curves. The resulting derivatives facilitate crafting equations for tangents and normals to the curves.

Executing Implicit Differentiation

With explicit functions delineated as y = ax+..., standard differentiation applies via the chain rule. Conversely, for equations implicitly stated as x + y = a, an adapted form of the chain rule is employed, predicated on the premise that y functions in terms of x.

  • Differentiate each term on both sides of the equation concerning x. For terms involving y, append a multiplication by dy/dx.
  • Subsequent to differentiation, we isolate dy/dx.

Consider the following circle equation. Derive its differential.

x^2 + y^2 = 49


Begin by differentiating each segment of the equation. Note, the y part necessitates a dy/dx multiplication.

2x + 2y(dy/dx) = 0

dy/dx = -2x/2y

Determine \(\frac{dy}{dx}\) given

\[x^5 + y^2 - x \cdot y = 9\]


\[\frac{d}{dx}(x^5) + \frac{d}{dx}(y^2) - \frac{d}{dx}(x \cdot y) = \frac{d}{dx}(9) = 0\]

In this instance, there's also a product of variables, necessitating inclusion in the differentiation process for the term xy. The formula to utilize is as follows.

\[\frac{d}{dx}(u \cdot v) = v \cdot \frac{d}{dx}(u) + u \cdot \frac{d}{dx}(v)\]

Thus, applying the formula to the xy term yields:

\[5x^4 + 2y \cdot \frac{dy}{dx} - (y + x \cdot \frac{dy}{dx}) = 0\]

\[5x^4 + 2y \cdot \frac{dy}{dx} - y - x \cdot \frac{dy}{dx} = 0\]

\[\frac{dy}{dx} \cdot (2y - x) = y - 5x^4\]

Isolating \(\frac{dy}{dx}\) gives:

\[\frac{dy}{dx} = \frac{y - 5x^4}{2y - x}\]

Advancing to Higher-Order Implicit Differentiation

For higher-order derivatives, the approach mirrors the initial step. The successive derivative is obtained by differentiating the preceding derivative, continuing for the required order. This principle is encapsulated by:

Here, n signifies the derivative's order.

\[\frac{d^n y}{dx^n} = \frac{d}{dx} (\frac{d^{n-1} y}{dx^{n-1}})\]

In differentiating implicit functions, the aforementioned strategy is applied to determine the initial lower-order derivative. To derive higher-order derivatives, the presented formula is utilized.

Derive the second-order differential of the expression:

\[x^2 + 2y = 3\]

Upon differentiating relative to x:

\[2x + 2 \cdot \frac{dy}{dx} = 0\]

\[\frac{dy}{dx} = -x\]

Further differentiation yields:

\[\frac{d^2 y}{dx^2} = -1\]

Deducing the Tangent Equation of a Curve via Implicit Differentiation

To deduce a tangent's equation, a linear representation, implicit differentiation first establishes the curve's slope. Provided the tangency point is identified, the linear equation is computed using the formula mentioned, with x1 and y1 representing the point's coordinates and a denoting the curve's derivative or slope.

\[y - y_1 = a(x - x_1)\]

A given curve's equation is:

\[x^3 + x^2 + x \cdot y = 3\]

Ascertain the tangent's equation at x = 1.

Identify the tangency point by substituting the given coordinate, deriving the respective y value.

\[1^3 + 1^2 + 1 \cdot y = 3 \Rightarrow y = 1\]

To formulate the tangent equation, also establish the curve's slope at the tangency.

\[\frac{d}{dx}(x^3) + \frac{d}{dx}(x^2) + \frac{d}{dx}(x \cdot y) = \frac{d}{dx} (3)\]

\(3x^2 + 2x + y + x \cdot \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{-3x^2 - 2x - y}{x} = \frac{-3 \cdot 1^2 - 2 \cdot 1 - 1}{1} = 6\)

The tangent equation is established using the provided formula.

\(y - y_1 = a(x - x_1) \Rightarrow y - 1 = 6(x - 1) \Rightarrow y = 6 \cdot x - 5\)

Formulating a Curve's Normal Equation Through Implicit Differentiation

The process to formulate a curve's normal equation mirrors that for a tangent, with the normal's slope being the negative reciprocal of the curve's tangent slope, as illustrated in the subsequent formula where N and T represent the slopes of the normal and tangent, respectively.

\(slope_N \cdot Slope_T = -1\)

Determine the normal slope for the curve showcased earlier.


Continuing from the previous case, obtaining the normal's slope utilizes the derivative ascertained.

\(\frac{dy}{dx} = -6\)

Applying the normal slope formula, given its perpendicularity to the curve, the normal's slope inverses to 1/6. Employing the straight line construction formula, the normal equation emerges as follows.

\(y - y_1 = \frac{dy}{dx} (x - x_1)\)

\(y - 1 = \frac{1}{6} (x - 1) \Rightarrow y = \frac{1}{6} \cdot x + \frac{5}{6}\)

Essential Insights on Implicit Differentiation

  • Implicit differentiation serves when equations involve both unknowns without segregation to any equation side.
  • All terms undergo differentiation, with dy/dx specifically applied to y terms.
  • For acquiring second or third derivatives, the initial derivative is differentiated per the required order.
  • The equations for tangents and normals to implicitly defined curves are ascertainable using their implicit derivatives and given coordinates.

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