Derivative Calculator

Quotient rule

Bottom times the derivative of the top, minus top times the derivative of the bottom, all over the bottom squared.

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

What it says

A quotient \(\frac{f}{g}\) is the product \(f \cdot g^{-1}\), so this rule is the product rule and the chain rule combined and then tidied. Working it out that way is a useful check that the minus sign belongs where it does.

The order in the numerator matters. \(f'g - fg'\) and \(fg' - f'g\) differ by a sign, and only the first is correct. A mnemonic that survives exams: low d-high minus high d-low, over low squared.

The result is undefined wherever \(g = 0\) — which is exactly where the original function is undefined too.

When it applies

  • One expression containing the variable is divided by another that also contains it.
  • Rational functions, and ratios such as \(\frac{\sin x}{x}\).
  • Not needed when the denominator is constant, or when the fraction can be split into simpler terms.

Five worked examples

Every line is the step the calculator would show, in the order it applies them. Each graph is live: hover it to read both curves and see the tangent whose slope is the derivative, drag to pan, scroll to zoom.

Example 1
\[\frac{d}{dx}\left[\frac{x}{x + 1}\right]\]

  1. Take \(f = x\), \(g = x + 1\).

    \[= \frac{1 \cdot \left(x + 1\right) - x \cdot 1}{\left(x + 1\right)^{2}}\]

  2. The numerator collapses.

    \[= \frac{1}{\left(x + 1\right)^{2}}\]
Answer
\[\frac{1}{\left(x + 1\right)^{2}}\]

A pole at \(x = -1\) splits the graph. f′ is positive everywhere it exists, so neither branch ever falls — the left one climbs from below, the right one approaches \(y = 1\) from underneath. f′ dies away as \(\left|x\right|\) grows, which is why f flattens.

Always positive, which fits: the function increases everywhere it is defined.

Open x/(x+1) in the calculator
Example 2
\[\frac{d}{dx}\left[\frac{x^{2} + 1}{x - 1}\right]\]

  1. Apply the rule with \(f = x^{2} + 1\), \(g = x - 1\).

    \[= \frac{2x\left(x - 1\right) - \left(x^{2} + 1\right)\cdot 1}{\left(x - 1\right)^{2}}\]

  2. Expand the numerator.

    \[= \frac{2x^{2} - 2x - x^{2} - 1}{\left(x - 1\right)^{2}}\]

  3. Collect like terms.

    \[= \frac{x^{2} - 2x - 1}{\left(x - 1\right)^{2}}\]
Answer
\[\frac{x^{2} - 2 x - 1}{\left(x - 1\right)^{2}}\]

Pole at \(x = 1\), with f′ vanishing at \(1 \pm \sqrt{2}\) — one turning point on each branch, a maximum on the left and a minimum on the right.

Expanding the numerator is worth the effort — it is where sign errors show up.

Open (x^2+1)/(x-1) in the calculator
Example 3
\[\frac{d}{dx}\left[\frac{\sin\left(x\right)}{x}\right]\]

  1. With \(f = \sin x\), \(g = x\).

    \[= \frac{\cos x \cdot x - \sin x \cdot 1}{x^{2}}\]

  2. Tidy.

    \[= \frac{x\cos x - \sin x}{x^{2}}\]
Answer
\[\frac{x \cos\left(x\right) - \sin\left(x\right)}{x^{2}}\]

The sinc curve, peaking near 1 at the origin and rippling away with shrinking amplitude, f′ crossing zero at every ripple.

The sinc function. Its derivative is zero exactly where \(\tan x = x\).

Open sin(x)/x in the calculator
Example 4
\[\frac{d}{dx}\left[\frac{e^{x}}{x^{2}}\right]\]

  1. With \(f = e^{x}\), \(g = x^{2}\).

    \[= \frac{e^{x}x^{2} - e^{x}\cdot 2x}{x^{4}}\]

  2. Cancel one factor of \(x\).

    \[= \frac{e^{x}\left(x - 2\right)}{x^{3}}\]
Answer
\[\frac{x e^{x} - 2 e^{x}}{x^{3}}\]

A pole at the origin. On the right f falls to a minimum at \(x = 2\), where f′ crosses zero, then climbs away.

Cancelling after applying the rule is normal; the derivative is zero at \(x = 2\), the minimum of the curve.

Open e^x/x^2 in the calculator
Example 5
\[\frac{d}{dx}\left[\frac{\sin\left(x\right)}{\cos\left(x\right)}\right]\]

  1. This quotient is \(\tan x\). Apply the rule.

    \[= \frac{\cos x\cos x - \sin x\left(-\sin x\right)}{\cos^{2}x}\]

  2. The numerator is \(\cos^{2}x + \sin^{2}x = 1\).

    \[= \frac{1}{\cos^{2}x} = \sec^{2}x\]
Answer
\[\sec^{2}\left(x\right)\]

This quotient is \(\tan x\). f′ never drops below 1, so every branch climbs and none of them flattens.

This is where the standard derivative of \(\tan\) comes from — it is not an independent fact.

Open sin(x)/cos(x) in the calculator

Watch out for

  • The numerator is \(f'g - fg'\), in that order. Reversing it flips the sign of the whole answer.
  • The denominator is \(g^{2}\), not \(g\) and not \(\left(g'\right)^{2}\).
  • A constant denominator does not need this rule: \(\frac{f}{c} = \frac{1}{c}f\).
Product ruleChain rulePower rule