Quotient Rule

The Quotient Rule is a differentiation rule used to determine the derivative of a function that is expressed as the quotient (division) of two differentiable functions.

If a function is written as y=\frac{u(x)}{v(x)}, where v(x) ≠ 0, then its derivative can be found using the Quotient Rule:

\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{v(x)u'(x)-u(x)v'(x)}{[v(x)]^2}\\[3pts]\text{or}\\[3pts]\left(\frac{u}{v}\right)' = \frac{vu'-uv'}{v^2}

where,

• u(x) is the first function that is differentiable, 
• u'(x) is the derivative of function u(x), 
• v(x) is the second function, which is a differentiable function, and
• v'(x) is the derivative of the function v(x).

Steps to Apply

Step 1: Write the individual functions as u(x) and v(x).

Step 2: Find the derivative of the individual function u(x) and v(x), i.e. find u'(x) and v'(x).

Step 3: Now apply the quotient rule formula: \frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{v(x)u'(x)-u(x)v'(x)}{[v(x)]^2}

Step 4: Simplify the expression.

Proof

We can derive the quotient rule using the following methods:

1. Using Chain Rule

Given, y=\frac{u(x)}{v(x)}

Using the quotient as a product, y=u(x)[v(x)]^{-1}

Applying the Product Rule, y' = u'(x)[v(x)]^{-1} + u(x)\frac{d}{dx}[v(x)]^{-1}

Using the Power Rule, \frac{d}{dx}[v(x)]^{-1} = -\frac{v'(x)}{[v(x)]^2}

Substituting, y' = \frac{u'(x)}{v(x)} - \frac{u(x)v'(x)}{[v(x)]^2}

Taking the common denominator v²(x)

y' = \frac{u'(x)v(x)-u(x)v'(x)} {[v(x)]^2}

Therefore, \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x)u'(x)-u(x)v'(x)} {[v(x)]^2}

Hence, the Quotient Rule is proved.

2. Using Derivative and Limit Properties

Let , f(x)=\frac{u(x)}{v(x)}, where u(x) and v(x) are differentiable functions and v(x) ≠ 0.

Using the definition of a derivative, f'(x)=\lim\limits_{h\to 0}\frac{f(x+h)-f(x)}{h}

Substituting f(x)=\dfrac{u(x)}{v(x)},

f'(x) = \lim\limits_{h\to 0} \frac{\frac{u(x+h)}{v(x+h)}-\frac{u(x)}{v(x)}}{h}\\[3pts]f'(x) = \lim\limits_{h\to 0} \frac{u(x+h)v(x)-u(x)v(x+h)} {h\,v(x+h)v(x)}

Add and subtract u(x)v(x) in the numerator: \lim_{h\to 0} \frac{ u(x+h)v(x)-u(x)v(x) + u(x)v(x)-u(x)v(x+h) } {h\,v(x+h)v(x)}

Factor the common terms: \lim\limits_{h\to 0} \frac{ v(x)\big[u(x+h)-u(x)\big] - u(x)\big[v(x+h)-v(x)\big] } {h\,v(x+h)v(x)}

Separate the expression into two terms: \lim\limits_{h\to 0} \left[ \frac{v(x)\big(u(x+h)-u(x)\big)} {h\,v(x+h)v(x)} - \frac{u(x)\big(v(x+h)-v(x)\big)} {h\,v(x+h)v(x)} \right]

Simplifying and applying the limit,

= \lim\limits_{h\to 0} \left[ \frac{u(x+h)-u(x)} {h\,v(x+h)} - \frac{u(x)\big(v(x+h)-v(x)\big)} {h\,v(x+h)v(x)} \right]= \frac{u'(x)}{v(x)} - \frac{u(x)v'(x)}{v^2(x)}

Taking the common denominator v²(x),

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

Therefore, \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x)u'(x)-u(x)v'(x)} {[v(x)]^2}

Hence, the Quotient Rule is proved.

Common Mistakes While Applying

1. Reversing the order of terms in the numerator

formula is:

\left(\frac{u}{v}\right)'=\frac{vu'-uv'}{v^2}

2. Forgetting to square the denominator : The denominator in the Quotient Rule is always v², not v.

\left(\frac{u}{v}\right)'=\frac{vu'-uv'}{v^2}

3. Forgetting the negative sign

The second term in the numerator is subtracted: vu′ − uv′.

4. Not considering where the denominator is zero :The Quotient Rule requires

v(x) ≠ 0.. A quotient function is not defined at points where its denominator is zero

Practice Questions on Quotient Rule