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Quotient Rule Calculator

Last updated: March 28, 2026

Calculator

Results

f(x) = ax + b

8

g(x) = cx + d

9

f'(x) = a

3

g'(x) = c

5

f(x)/g(x)

0.888889

f'g − fg' (numerator)

-13

g(x)² (denominator)

81

(f/g)' = (f'g − fg')/g²

-0.160494

Results

f(x) = ax + b

8

g(x) = cx + d

9

f'(x) = a

3

g'(x) = c

5

f(x)/g(x)

0.888889

f'g − fg' (numerator)

-13

g(x)² (denominator)

81

(f/g)' = (f'g − fg')/g²

-0.160494

The Quotient Rule Calculator computes the derivative of a quotient of two linear functions using the quotient rule of calculus. Given $$f(x) = ax + b$$ and $$g(x) = cx + d$$, this tool evaluates the quotient $$f/g$$, applies the quotient rule formula $$\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$$, and breaks down every component of the calculation at your chosen x-value. It is an essential study and verification tool for anyone learning or applying differentiation.

The quotient rule is the differentiation rule for ratios of functions. While students often try to differentiate a quotient by separately differentiating the numerator and denominator, this approach is incorrect. The correct formula accounts for the interplay between the changing numerator and the changing denominator:

$$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x) \cdot g(x) - f(x) \cdot g'(x)}{[g(x)]^2}$$

A popular mnemonic is "low d-high minus high d-low, over the square of what's below," where "high" is the numerator $$f$$, "low" is the denominator $$g$$, and "d" means "derivative of." The quotient rule can be derived from the product rule by writing $$f/g = f \cdot g^{-1}$$ and applying the product rule together with the chain rule for $$g^{-1}$$.

For linear functions $$f(x) = ax + b$$ and $$g(x) = cx + d$$, the quotient rule simplifies elegantly. Since $$f' = a$$ and $$g' = c$$, the numerator of the derivative becomes $$a(cx + d) - c(ax + b) = ad - bc$$, a constant independent of $$x$$. The full derivative is therefore $$\frac{ad - bc}{(cx + d)^2}$$. Notice that the numerator $$ad - bc$$ has the same structure as a 2×2 determinant — this is not a coincidence and connects to deeper results in linear algebra.

The quotient rule is indispensable throughout mathematics and its applications. In physics, velocity is the derivative of position, and when position involves a ratio (such as in certain projectile or planetary motion problems), the quotient rule is required. In economics, average cost is total cost divided by quantity, and its derivative (marginal average cost) requires the quotient rule. In probability, Bayes' theorem involves ratios of probabilities whose derivatives (with respect to parameters) use the quotient rule.

An important caveat: the quotient rule is undefined wherever $$g(x) = 0$$, because division by zero is undefined. For the linear denominator $$g(x) = cx + d$$, this occurs at $$x = -d/c$$ (when $$c \neq 0$$). At this point, the original function $$f/g$$ has a vertical asymptote, and neither the function nor its derivative exists. This calculator returns NaN (not a number) if you evaluate at a point where the denominator is zero.

Enter the coefficients of two linear functions and an x-value to see the quotient rule applied step by step, with every intermediate quantity displayed for complete transparency.

Visual Analysis

How It Works

The Quotient Rule Calculator evaluates the derivative of a ratio of two linear functions at a specified point.

Step 1: Define the functions.

$$f(x) = ax + b, \quad g(x) = cx + d$$

Step 2: Compute derivatives.

$$f'(x) = a, \quad g'(x) = c$$

Step 3: Evaluate at x₀.

$$f(x_0) = ax_0 + b, \quad g(x_0) = cx_0 + d$$

Step 4: Apply the quotient rule.

$$\left(\frac{f}{g}\right)'(x_0) = \frac{f'(x_0) \cdot g(x_0) - f(x_0) \cdot g'(x_0)}{[g(x_0)]^2}$$

$$= \frac{a(cx_0 + d) - c(ax_0 + b)}{(cx_0 + d)^2}$$

$$= \frac{ad - bc}{(cx_0 + d)^2}$$

The numerator $$ad - bc$$ is constant for all x — this is a special property of the quotient of two linear functions. The denominator $$(cx_0 + d)^2$$ is always positive (when defined), so the sign of the derivative depends entirely on the sign of $$ad - bc$$.

Understanding Your Results

The f(x) and g(x) values show the numerator and denominator functions evaluated at your point. If g(x) = 0, the quotient and its derivative are undefined.

The f'(x) and g'(x) values are the constant derivatives of the linear functions (simply the coefficients a and c).

The f(x)/g(x) value is the quotient at the evaluation point — the ratio you are differentiating.

The f'g − fg' numerator shows the cross-difference that appears in the quotient rule. For linear functions, this simplifies to the constant $$ad - bc$$. If this is zero, the derivative is zero everywhere (the quotient is constant).

The g(x)² denominator is always positive when the denominator function is nonzero, ensuring the derivative has a well-defined sign determined by the numerator.

The (f/g)' value is the final derivative of the quotient at your evaluation point. It represents the instantaneous rate of change of the ratio.

Worked Examples

f(x) = 3x + 2, g(x) = 5x − 1 at x = 2

Inputs

a3
b2
c coeff5
d-1
x02

Results

fVal8
gVal9
fPrime3
gPrime5
quotient0.888889
numerator-13
denominator81
quotientDeriv-0.160494

f(2) = 8, g(2) = 9. f'g − fg' = 3(9) − 8(5) = 27 − 40 = −13. g² = 81. (f/g)' = −13/81 ≈ −0.1605. Check: ad − bc = 3(−1) − 2(5) = −3 − 10 = −13. ✓

f(x) = 4x + 6, g(x) = 2x + 3 at x = 1

Inputs

a4
b6
c coeff2
d3
x01

Results

fVal10
gVal5
fPrime4
gPrime2
quotient2
numerator0
denominator25
quotientDeriv0

f(1) = 10, g(1) = 5. ad − bc = 4(3) − 6(2) = 12 − 12 = 0. The derivative is zero because f = 2g (the ratio is constant at 2 for all x where g ≠ 0). This illustrates that proportional functions have a zero quotient derivative.

Frequently Asked Questions

The quotient rule is a formula for differentiating a ratio of two functions: $$\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$$. It is necessary because the derivative of a quotient is not simply the quotient of the derivatives. The rule accounts for both the numerator and denominator changing simultaneously.

When $$g(x_0) = 0$$, the quotient $$f/g$$ is undefined (division by zero), and the derivative does not exist at that point. For the linear denominator $$cx + d$$, this occurs at $$x = -d/c$$. The function has a vertical asymptote there, and the calculator returns NaN.

For linear functions, the quotient rule numerator $$ad - bc$$ is indeed the determinant of the matrix $$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$. This is because the quotient of two linear functions is a Möbius transformation, and its derivative involves this determinant. If $$ad - bc = 0$$, the two functions are proportional and their ratio is constant.

Yes. You can rewrite $$f/g = f \cdot g^{-1}$$ and apply the product rule combined with the chain rule: $$(f \cdot g^{-1})' = f' \cdot g^{-1} + f \cdot (-g^{-2} \cdot g')$$. Simplifying gives the same quotient rule formula. Many mathematicians prefer this approach since it reduces the number of rules to memorize.

For linear functions, the derivative $$\frac{ad - bc}{(cx_0 + d)^2}$$ has the same sign as $$ad - bc$$ (since the denominator is always positive). If $$ad > bc$$, the quotient is increasing; if $$ad < bc$$, it is decreasing; if $$ad = bc$$, the quotient is constant.

In economics, average cost is $$\bar{C}(q) = C(q)/q$$. To find whether average cost is increasing or decreasing, you differentiate using the quotient rule: $$\bar{C}'(q) = \frac{C'(q) \cdot q - C(q)}{q^2}$$. Average cost decreases when marginal cost $$C'(q)$$ is below average cost, a key insight for production decisions.

Sources & Methodology

Stewart, James. Calculus: Early Transcendentals, 8th ed., Cengage Learning, 2015. Thomas, George B. et al. Thomas' Calculus, 14th ed., Pearson, 2017. Spivak, Michael. Calculus, 4th ed., Publish or Perish, 2008. Weisstein, Eric W. "Quotient Rule." MathWorld — A Wolfram Web Resource.
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