IB Applications and Interpretation SL · Differential and integral calculus · Section 3 of 7
Differentiation rules
3 worked examples · By Álvaro López
Key idea
Differentiating means applying a few rules to each term of the function. They cover polynomials and expressions with powers, including negative and fractional exponents, which are the ones that appear most often in the exam.
Basic derivatives
The power rule
There is a general rule for powers of x that works for any exponent.
Constants, sums and differences
Negative and fractional exponents
In the exam you will see functions such as x21 or x. Before differentiating, write them as powers of x: x21=x−2, x1=x−1 and x=x1/2. This way the power rule works without any change.
Expanding before differentiating
For this topic, expanding is all you need. If the function is a product of a monomial and a polynomial, such as x(3x2−1), expand it first to get 3x3−x and differentiate afterwards. This way you do not need to remember any extra rule, and the result is easier to check.
Worked examples
Example 1★ Basic
A polynomial of degree four
Let f(x)=6x4−5x2+3x−9.
a) Find f′(x).
b) Find f′(1).
Solution
a) We differentiate term by term:
6x4→6⋅4x3=24x3
−5x2→−5⋅2x=−10x
3x→3
−9→0
Therefore f′(x)=24x3−10x+3.
b) Substitute x=1: f′(1)=24−10+3=17.
Example 2★★ Medium
Expanding and negative exponents
Let g(x)=x(3x2−1) and h(x)=x26−x1, with x=0.
a) Find g′(x).
b) Find h′(x) and write it without negative exponents.
c) Calculate h′(2).
Solution
a) Expand first: g(x)=3x3−x. Then
g′(x)=9x2−1
b) We write h(x)=6x−2−x−1. We differentiate each term:
6x−2→6⋅(−2)x−3=−12x−3−x−1→−(−1)x−2=x−2
So
h′(x)=−12x−3+x−2=−x312+x21
c)h′(2)=−812+41=−23+41=−45.
Example 3★★★ Hard
Two constants and two conditions
The curve y=px3+qx, where p and q are constants, passes through the point (1,4). At x=1, its gradient is 7.
a) Find p and q.
b) Using the values of p and q from the previous part, find dxdy.
Solution
a) The curve passing through (1,4) means p+q=4. The derivative is
dxdy=3px2+q
and the gradient at x=1 is 3p+q=7. We have the system
{p+q3p+q=4=7
Subtracting the first equation from the second, 2p=3, so p=23, and then q=4−23=25.
Check: 3⋅23+25=29+25=7.
b) Substituting,
dxdy=3⋅23x2+25=29x2+25
In the exam: two constants need two equations, and each condition gives one equation: the point gives an equation using the function, and the gradient gives another using the derivative. If you write only one, you cannot find both values.
Common mistakes
Summary
The derivative of a constant is 0, and the derivative of x is 1.
Power rule: if f(x)=xn, then f′(x)=nxn−1, for any real exponent.
Constants come outside, and the derivative is applied term by term across sums and differences.
Write xm1 as x−m and x as x1/2 before differentiating.
If a product is simple, expand it before differentiating.
