How to Do Derivatives of Fractions
Learning how to do derivatives of fractions is essential for calculus, especially when the numerator, denominator, or both contain variables. The main tool for solving these problems is the quotient rule, which allows you to differentiate a fraction without dividing the derivative of the numerator by the derivative of the denominator. By understanding the rule, working through several examples, and learning how to avoid common mistakes, you can calculate derivatives of fractions confidently and accurately.
What Is a Derivative of a Fraction?
A fraction is generally written in the form
[ f(x)=\frac{u(x)}{v(x)}, ]
where:
- (u(x)) is the numerator
- (v(x)) is the denominator
- (v(x)\neq 0)
To find the derivative (f'(x)), use the quotient rule:
[ \left(\frac{u}{v}\right)'= \frac{v,u'-u,v'}{v^2} ]
In words, the derivative of a quotient equals:
The denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all divided by the square of the denominator.
The order of the terms is the kind of thing that makes a real difference. The expression (vu'-uv') is not the same as (uv'-vu').
Step-by-Step Method
To differentiate a fraction, follow these steps:
- Identify the numerator and denominator.
- Find the derivative of the numerator.
- Find the derivative of the denominator.
- Apply the quotient rule.
- Substitute the derivatives into the formula.
- Simplify the result when possible.
Suppose you are given:
[ y=\frac{x^2+1}{x-3} ]
The numerator is:
[ u=x^2+1 ]
The denominator is:
[ v=x-3 ]
Their derivatives are:
[ u'=2x ]
and
[ v'=1 ]
Now apply the quotient rule:
[ y'=\frac{(x-3)(2x)-(x^2+1)(1)}{(x-3)^2} ]
Expand the numerator:
[ y'=\frac{2x^2-6x-x^2-1}{(x-3)^2} ]
Combine like terms:
[ y'=\frac{x^2-6x-1}{(x-3)^2} ]
Because of this, the final answer is:
[ \boxed{y'=\frac{x^2-6x-1}{(x-3)^2}} ]
Example: Differentiating a Rational Function
Consider the function:
[ y=\frac{3x^2+2}{x^2+1} ]
Let:
[ u=3x^2+2 ]
and
[ v=x^2+1 ]
Differentiate each part:
[ u'=6x ]
[ v'=2x ]
Apply the quotient rule:
[ y'=\frac{(x^2+1)(6x)-(3x^2+2)(2x)}{(x^2+1)^2} ]
Expand the terms in the numerator:
[ y'=\frac{6x^3+6x-6x^3-4x}{(x^2+1)^2} ]
The (6x^3) terms cancel:
[ y'=\frac{2x}{(x^2+1)^2} ]
Thus:
[ \boxed{y'=\frac{2x}{(x^2+1)^2}} ]
This example demonstrates an important principle: simplifying after applying the quotient rule can reveal cancellations that make the answer much easier to understand.
Example: A Fraction with Trigonometric Functions
Let:
[ y=\frac{\sin x}{x} ]
The numerator is (u=\sin x), so:
[ u'=\cos x ]
The denominator is (v=x), so:
[ v'=1 ]
Using the quotient rule:
[ y'=\frac{x\cos x-\sin x(1)}{x^2} ]
Therefore:
[ \boxed{y'=\frac{x\cos x-\sin x}{x^2}} ]
The quotient rule works in the same way when the numerator and denominator contain trigonometric, exponential, logarithmic, or other functions. You simply differentiate each part using the appropriate rule Most people skip this — try not to..
Example: Differentiating a Fraction Raised to a Power
Suppose:
[ y=\left(\frac{x}{x+1}\right)^3 ]
This problem requires both the chain rule and the quotient rule Most people skip this — try not to..
First, differentiate the expression inside the parentheses. For
[ u=\frac{x}{x+1}, ]
the quotient rule gives:
[ u'=\frac{(x+1)(1)-x(1)}{(x+1)^2} ]
Simplify:
[ u'=\frac{x+1-x}{(x+1)^2} ]
[ u'=\frac{1}{(x+1)^2} ]
Now apply the chain rule:
[ y'=3\left(\frac{x}{x+1}\right)^2\cdot\frac{1}{(x+1)^2} ]
Combine the denominators:
[ \boxed{y'=\frac{3x^2}{(x+1)^4}} ]
A useful general formula is:
[ \left(\frac{u}{v}\right)^n ]
differentiates as:
[ n\left(\frac{u}{v}\right)^{n-1} \left(\frac{vu'-uv'}{v^2}\right) ]
Remember that the outer power does not disappear. The factor (n) comes from differentiating the exponent.
The Reciprocal Rule
Sometimes the numerator is simply (1). For example:
[ y=\frac{
