Difficulty: Intermediate | Prerequisites: Algebra, exponent rules, basic limit definition of derivative
Derivative computation is the core mechanical skill of MAT 210. Every applied problem in the course, from finding critical points to computing rates of change, starts with taking a derivative. This topic covers the rules you need to differentiate any function you will see on the exam: power rule, product rule, quotient rule, chain rule, and the specific formulas for exponentials and logarithms. If you cannot differentiate fluently, every subsequent topic will feel harder than it needs to.
The product rule handles f(x) = u(x) · v(x). The quotient rule handles f(x) = u(x)/v(x). The chain rule handles compositions like f(g(x)). For exponentials, the derivative of e^(u) is e^(u) · u'. For logarithms, the derivative of ln(u) is u'/u. Most exam problems layer two or three of these rules together.
Product rule
If f(x) = u(x) · v(x), then f'(x) = u'(x) · v(x) + u(x) · v'(x). Think of it as: "derivative of the first times the second, plus the first times the derivative of the second."
Quotient rule
If f(x) = u(x) / v(x), then f'(x) = [u'(x) · v(x) − u(x) · v'(x)] / [v(x)]². In simple terms, "low d-high minus high d-low, over the square of what's below."
Chain rule
If f(x) = g(h(x)), then f'(x) = g'(h(x)) · h'(x). In simple terms, you differentiate the outer function (leaving the inner function alone) and multiply by the derivative of the inner function.
General power rule
A special case of the chain rule: if f(x) = [u(x)]^n, then f'(x) = n · [u(x)]^(n−1) · u'(x). This applies whether n is positive, negative, or fractional.
Exponential derivative
The derivative of e^(u(x)) is e^(u(x)) · u'(x). The exponential function reproduces itself, then you multiply by the derivative of the exponent.
Logarithmic derivative
The derivative of ln(u(x)) is u'(x) / u(x). In simple terms, take the derivative of what is inside the ln and divide by what is inside the ln.
Worked example: h(x) = (5x² − 6√x) / e^(2x³)
Rewrite 6√x as 6x^(1/2). Let u = 5x² − 6x^(1/2) and v = e^(2x³).
u' = 10x − 3x^(−1/2)
v' = e^(2x³) · 6x² (chain rule on the exponent)
h'(x) = [(10x − 3x^(−1/2)) · e^(2x³) − (5x² − 6x^(1/2)) · e^(2x³) · 6x²] / [e^(2x³)]²
You do not need to simplify further on this exam.
Worked example: g(x) = 3(2x⁴ − 6/x)^(−8)
Rewrite 6/x as 6x^(−1). Let u = 2x⁴ − 6x^(−1). Apply the general power rule:
g'(x) = 3 · (−8) · (2x⁴ − 6x^(−1))^(−9) · u'
u' = 8x³ + 6x^(−2)
g'(x) = −24(2x⁴ − 6x^(−1))^(−9) · (8x³ + 6x^(−2))
Worked example: k(x) = ln(2x⁵ − 10x + 3)
Let u = 2x⁵ − 10x + 3.
k'(x) = u'/u = (10x⁴ − 10) / (2x⁵ − 10x + 3)
Worked example: m(x) = (3/(2x⁵)) · √(x² + 1)
Rewrite as m(x) = (3/2)x^(−5) · (x² + 1)^(1/2). Let u = (3/2)x^(−5) and v = (x² + 1)^(1/2).
u' = (3/2)(−5)x^(−6) = −(15/2)x^(−6)
v' = (1/2)(x² + 1)^(−1/2) · 2x = x(x² + 1)^(−1/2)
m'(x) = u' · v + u · v' = −(15/2)x^(−6) · (x² + 1)^(1/2) + (3/2)x^(−5) · x(x² + 1)^(−1/2)
Worked example: f(x) = (7√x + 3/x²)³ · ln(x³ + 2x)
Let u = (7x^(1/2) + 3x^(−2))³ and v = ln(x³ + 2x).
u' requires the chain rule: 3(7x^(1/2) + 3x^(−2))² · (7 · (1/2)x^(−1/2) + 3(−2)x^(−3))
u' = 3(7x^(1/2) + 3x^(−2))² · ((7/2)x^(−1/2) − 6x^(−3))
v' = (3x² + 2) / (x³ + 2x)
f'(x) = u' · v + u · v'
Worked example: A(r) = 1000(1 + r/12)^60
Differentiating with respect to r using the chain rule:
A'(r) = 1000 · 60 · (1 + r/12)^59 · (1/12)
A'(r) = 5000(1 + r/12)^59
At r = 0.0395: A'(0.0395) = 5000(1 + 0.0395/12)^59. This gives the rate of change of the account balance per unit change in the interest rate, measured in dollars per unit of r.
Worked example: V(t) = t³ · e^(0.02t²)
Let u = t³ and v = e^(0.02t²).
u' = 3t²
v' = e^(0.02t²) · 0.04t
V'(t) = 3t² · e^(0.02t²) + t³ · e^(0.02t²) · 0.04t
V'(t) = e^(0.02t²)(3t² + 0.04t⁴)
At t = 5: V'(5) = e^(0.02 · 25)(3 · 25 + 0.04 · 625) = e^(0.5)(75 + 25) = 100e^(0.5) ≈ 164.87 litres per hour.
\text{Power rule: } \frac{d}{dx}[x^n] = nx^{n-1}\text{Product rule: } \frac{d}{dx}[u \cdot v] = u'v + uv'\text{Quotient rule: } \frac{d}{dx}\left[\frac{u}{v}\right] = \frac{u'v - uv'}{v^2}\text{Chain rule: } \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\text{General power rule: } \frac{d}{dx}[u^n] = n \cdot u^{n-1} \cdot u'\text{Exponential: } \frac{d}{dx}[e^{u}] = e^{u} \cdot u'\text{Logarithm: } \frac{d}{dx}[\ln(u)] = \frac{u'}{u}Derivatives measure instantaneous rates of change. In physics, the derivative of position gives velocity. In economics, the derivative of a cost function gives marginal cost, the extra cost of producing one more unit. In chemistry, reaction rates are derivatives of concentration with respect to time.
Students often forget to apply the chain rule when differentiating composite functions. If there is a function inside another function, you need the chain rule. The derivative of e^(2x) is 2e^(2x), not e^(2x).
Students sometimes apply the product rule when they should use the chain rule, and vice versa. Product rule: two separate functions multiplied together. Chain rule: one function plugged into another.
When using the quotient rule, students frequently get the numerator subtraction backwards. It is u'v minus uv', not the other way around.
Students often forget to rewrite roots and fractions as powers before differentiating. √x = x^(1/2), 1/x² = x^(−2), and so on. This makes the power rule and chain rule much easier to apply.
⚠️ Problem 9 on the sample exam is worth 15 points and asks for five separate derivatives. Each one layers multiple rules together. Practise identifying which rules to use before you start computing.
⚠️ Problems 11 and 12 ask you to compute a derivative and then evaluate it at a specific value. Do not forget units in your final answer.
⚠️ Rewriting expressions before differentiating (roots as fractional powers, fractions as negative powers) is worth doing every time. It reduces errors.
True or False: The derivative of e^(3x) is e^(3x). Answer: False. It is 3e^(3x). You must apply the chain rule.
Fill in the blank: The derivative of ln(u) is ______. Answer: u'/u.
True or False: To differentiate f(x) = x² · ln(x), you use the chain rule. Answer: False. You use the product rule (two separate functions multiplied).
Fill in the blank: Before differentiating √x, rewrite it as ______. Answer: x^(1/2).
True or False: In the quotient rule, the numerator is u'v + uv'. Answer: False. It is u'v − uv' (subtraction, not addition).
Q: Find the derivative of k(x) = ln(2x⁵ − 10x + 3).
A: k'(x) = (10x⁴ − 10) / (2x⁵ − 10x + 3). Apply the logarithmic derivative rule: derivative of the inside over the inside.
Q: Find V'(t) for V(t) = t³ · e^(0.02t²).
A: V'(t) = 3t² · e^(0.02t²) + t³ · e^(0.02t²) · 0.04t = e^(0.02t²)(3t² + 0.04t⁴). Product rule, with the chain rule on the exponential factor.
Q: Evaluate V'(5) for the function above. Include units if V is in litres and t in hours.
A: V'(5) = e^(0.5)(75 + 25) = 100e^(0.5) ≈ 164.87 litres per hour.
Q: Find A'(r) for A(r) = 1000(1 + r/12)^60.
A: A'(r) = 1000 · 60 · (1 + r/12)^59 · (1/12) = 5000(1 + r/12)^59. Chain rule with the inner function being (1 + r/12).
Q: Differentiate g(x) = 3(2x⁴ − 6/x)^(−8).
A: g'(x) = −24(2x⁴ − 6x^(−1))^(−9) · (8x³ + 6x^(−2)). General power rule (chain rule with a power).
Derivative computation feeds directly into critical points (set the derivative to zero and solve), optimisation (find the price or quantity that maximises revenue), and rates of change (evaluate the derivative at a specific input). The chain rule in particular is essential for any applied problem involving exponential growth, compound interest, or logarithmic models.
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