Derivative Calculator

The Quotient Rule: Differentiation Made Easy

The quotient rule is used when a function is defined as one function divided by another. The principle is: Differentiate the numerator, multiply by the denominator, then subtract the numerator times the derivative of the denominator — and divide everything by the square of the denominator.

📜 Formula of the Quotient Rule

If a function $f(x)$ is the quotient of two functions $u(x)$ and $v(x)$, the formula is:

Given the function:

$$ f(x) = \frac{u(x)}{v(x)} $$

Then its derivative is:

$$ f'(x) = \frac{u'(x)\,v(x) - u(x)\,v'(x)}{\big(v(x)\big)^2} $$

💡 Good to know: You can think of the quotient rule as a structured version of the product rule with $v(x)^{-1}$, but the explicit formula above is usually the safest to apply. Also, the denominator $v(x)$ must not be zero on the domain where you apply the rule.

🪜 Application in 3 Simple Steps

Use the quotient rule whenever your function is a fraction of two differentiable functions.

  1. Identify numerator and denominator: Set $u(x)$ to the numerator and $v(x)$ to the denominator.
  2. Differentiate separately: Compute $u'(x)$ and $v'(x)$ using the known rules (power rule, chain rule, etc.).
  3. Apply the quotient rule formula: Insert into $f'(x) = \dfrac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$ and (optionally) simplify.

✏️ Examples for Illustration

The quotient rule becomes intuitive with concrete examples.

Example 1: Polynomial divided by Polynomial

Given the function: $f(x) = \dfrac{x^2 + 1}{x - 1}$

  • 1. Identify: $u(x) = x^2 + 1$, $v(x) = x - 1$.
  • 2. Differentiate: $u'(x) = 2x$, $v'(x) = 1$.
  • 3. Apply the formula: $$ f'(x) = \frac{(2x)(x-1) - (x^2+1)(1)}{(x-1)^2} = \frac{2x^2 - 2x - x^2 - 1}{(x-1)^2} = \frac{x^2 - 2x - 1}{(x-1)^2}. $$

So the result is: $f'(x) = \dfrac{x^2 - 2x - 1}{(x-1)^2}$

Example 2: Trigonometric over Polynomial

Given the function: $f(x) = \dfrac{\sin(x)}{x}$

  • 1. Identify: $u(x) = \sin(x)$, $v(x) = x$.
  • 2. Differentiate: $u'(x) = \cos(x)$, $v'(x) = 1$.
  • 3. Apply the formula: $$ f'(x) = \frac{\cos(x)\cdot x - \sin(x)\cdot 1}{x^2} = \frac{x\cos(x) - \sin(x)}{x^2}. $$

So the result is: $f'(x) = \dfrac{x\cos(x) - \sin(x)}{x^2}$

Example 3: Exponential over Polynomial

Given the function: $f(x) = \dfrac{e^x}{x^2}$

  • 1. Identify: $u(x) = e^x$, $v(x) = x^2$.
  • 2. Differentiate: $u'(x) = e^x$, $v'(x) = 2x$.
  • 3. Apply the formula: $$ f'(x) = \frac{e^x\cdot x^2 - e^x \cdot 2x}{(x^2)^2} = \frac{e^x(x^2 - 2x)}{x^4} = \frac{e^x(x-2)}{x^3}\quad (\text{optional simplification}). $$

So the result is: $f'(x) = \dfrac{e^x(x^2 - 2x)}{x^4} = \dfrac{e^x(x-2)}{x^3}$