Quotient Rule for Differentiation

Neetesh Kumar | February 08, 2025 $\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space$ Share this Page on:

• 1. Definition of Quotient Rule
• 2. Explanation & Intuition
• 3. Solved Examples
• 4. Conclusion

Quotient Rule:
If $u(x)$ and $v(x)$ are differentiable functions, then the derivative of their quotient is given by:

$\boxed{\dfrac{d}{dx} \left[ \frac{u(x)}{v(x)} \right] = \frac{v(x). \dfrac{d}{dx} \bigg[u(x)\bigg] - u(x). \dfrac{d}{dx} \bigg[v(x)\bigg]}{\bigg[v(x)\bigg]^2} }$

This means that when differentiating a fraction, we take the derivative of the numerator, multiply it by the denominator, then subtract the numerator times the derivative of the denominator, and divide by the square of the denominator.

Example 1: Differentiating a Rational Function

Differentiate $f(x) = \dfrac{x^2}{x+1}$.

Solution:

Using the quotient rule, let:

• $u(x) = x^2$, so $u'(x) = 2x$.
• $v(x) = x+1$, so $v'(x) = 1$.

Applying the formula:

$f'(x) = \dfrac{(x+1)(2x) - (x^2)(1)}{(x+1)^2}$

$f'(x) = \dfrac{2x(x+1) - x^2}{(x+1)^2}$

$f'(x) = \dfrac{2x^2 + 2x - x^2}{(x+1)^2}$

$f'(x) = \dfrac{x^2 + 2x}{(x+1)^2}$

Thus, the final answer is: $\boxed{\frac{(x^2 + 2x)}{(x+1)^2}}$.

Example 2: Differentiating a Trigonometric Function

Differentiate $f(x) = \dfrac{\sin x}{x}$.

Solution:

Using the quotient rule, let:

• $u(x) = \sin x$, so $u'(x) = \cos x$.
• $v(x) = x$, so $v'(x) = 1$.

Applying the formula:

$f'(x) = \dfrac{x \cos x - \sin x}{x^2}$

Thus, the final answer is: $\boxed{\dfrac{x \cos x - \sin x}{x^2}}$.

Example 3: Finding the Rate of Change in Electrical Resistance

Ohm’s Law states that electrical resistance $R$ is given by: $R = \dfrac{V}{I}$

where $V$ is voltage and $I$ is current. Find the rate of change of resistance with respect to time when:

• $V = 10V$, $I = 5A$
• $\dfrac{dV}{dt} = 1 \frac{V}{s}$, $\dfrac{dI}{dt} = 0.5 \frac{A}{s}$.

Solution:

Using the quotient rule, let:

• $u = V$, so $u' = \dfrac{dV}{dt} = 1$.

• $v = I$, so $v' = \dfrac{dI}{dt} = 0.5$.

Applying the formula:

$\dfrac{dR}{dt} = \dfrac{I(1) - V(0.5)}{I^2}$

Substituting values:

$\dfrac{dR}{dt} = \dfrac{(5)(1) - (10)(0.5)}{(5)^2} = \dfrac{5 - 5}{25} = \dfrac{0}{25} = 0$

Thus, the resistance does not change at this moment: $\boxed{0}$.