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Longhand

Chapter 2: Derivatives

MATH 100 derivatives practice

The differentiation rules are short enough to write on one line each. What makes them difficult in practice is structural: deciding which rule the expression in front of you actually needs, and then keeping track of the order in which they nest when a function is a product of a quotient.

Almost all lost marks in this topic are bookkeeping rather than calculus. A dropped inner derivative, a reversed numerator in the quotient rule, or a substitution made before differentiating rather than after will each turn a correct method into a wrong answer.

These questions work through the power, sum, product and quotient rules individually and then in combination, and use them for what they are for: tangent lines, rates of change and second derivatives. The chain rule, exponentials, logarithms and trigonometric functions each have a topic of their own that follows this one.

Begin practice sessionBrowse all 10 questions

What you should be able to do

  • Apply the power rule to any exponent

    Including negative and fractional powers, which is why rewriting a root or a reciprocal as a power before differentiating is usually the first move.

  • Use the product rule without dropping a term

    The derivative of a product is u′v + uv′. It is never the product of the derivatives, and both cross terms are needed every time.

  • Get the quotient rule the right way round

    The numerator is u′v − uv′, starting with the derivative of the top. Reversing it produces exactly the negative of the correct answer, which is easy to miss.

  • Read a tangent line off a derivative

    The slope at a point is the derivative there, and the line through the point with that slope is the tangent. Finding where a tangent passes through a given point, or is shared by two curves, is the same idea run backwards.

  • Recognise a limit as a derivative

    A limit of the form (f(x) − f(a))/(x − a) is f′(a) in disguise. Spotting that turns an ugly limit into one line of the power rule.

  • Differentiate first, substitute second

    A derivative at a point is the derivative function evaluated there. Substituting the point early turns your function into a constant and the derivative into zero.

Where marks are usually lost

  • Multiplying the derivatives of a product

    Writing (uv)′ as u′v′ is the single most common error in the topic. Expanding the product first and differentiating term by term is a good way to check yourself.

  • Differentiating a quotient by differentiating top and bottom

    The derivative of u/v is not u′/v′. If the denominator is a single power of x, rewriting the quotient as a product with a negative exponent is often the cleanest route.

  • Reversing the quotient rule numerator

    uv′ − u′v gives the negative of the right answer. If your result has the correct shape but the wrong sign, check this first.

  • Confusing speed with velocity

    Velocity is the derivative of position and carries a sign; speed is its size. An object is speeding up when velocity and acceleration have the same sign, not simply when acceleration is positive.

  • Evaluating before differentiating

    f(1) and f′(1) are different questions. Differentiate the whole function, then substitute.

Practice questions

Every question below is original material written for Longhand. Open a solution to read the full working, or answer it yourself to have it counted towards your progress.

  • Introductory1 mark · 2 min

    Given f(x)=3x4−5x2+7f(x) = 3x^{4} - 5x^{2} + 7, find f′(1)f'(1).

    Answer this question yourself
  • Introductory2 marks · 3 min

    A student differentiates f(x)=5xx2+1f(x) = \dfrac{5x}{x^{2} + 1} as shown below. Which line contains the first error?

    Line 1:f′(x)=5 (x2+1)+5x⋅2x(x2+1)2Line 2:=5x2+5+10x2(x2+1)2Line 3:=5 (3x2+1)(x2+1)2\begin{aligned} \text{Line 1:}\quad f'(x) &= \frac{5\,(x^{2} + 1) + 5x\cdot 2x}{(x^{2} + 1)^{2}} \\[6pt] \text{Line 2:}\quad &= \frac{5x^{2} + 5 + 10x^{2}}{(x^{2} + 1)^{2}} \\[6pt] \text{Line 3:}\quad &= \frac{5\,(3x^{2} + 1)}{(x^{2} + 1)^{2}} \end{aligned}
    Answer this question yourself
  • Introductory2 marks · 3 min

    Find the yy-intercept of the tangent line to y=2x3−xy = 2x^{3} - x at the point where x=1x = 1.

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  • Standard2 marks · 3 min

    Let h(x)=(2x+1)(x2−3)h(x) = (2x + 1)(x^{2} - 3). Find h′(2)h'(2).

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  • Standard2 marks · 3 min

    Let q(x)=x2+1x−2q(x) = \dfrac{x^{2} + 1}{x - 2}. Which expression is q′(x)q'(x)?

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  • Standard3 marks · 4 min

    Let f(x)=3x2+52−xf(x) = \dfrac{3x^{2} + 5}{2 - x}. Find f′(1)f'(1).

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  • Standard2 marks · 3 min

    A particle moves along the xx-axis with position x(t)=t3−6t2+9tx(t) = t^{3} - 6t^{2} + 9t at time tt. Find its acceleration at t=1t = 1.

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  • Challenging3 marks · 5 min

    Evaluate the limit by recognising it as a derivative.

    lim⁡x→3x7−37x−3\lim_{x \to 3} \frac{x^{7} - 3^{7}}{x - 3}
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  • Challenging4 marks · 7 min

    Two distinct lines pass through the point (1,−6)(1, -6) and are tangent to the parabola y=2x2y = 2x^{2}. Find the slope of the steeper one.

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  • Capstone5 marks · 10 min

    There is exactly one line that is tangent to both of the parabolas below, at different points. Find its slope.

    y=x2andy=x2−4x+5y = x^{2} \qquad\text{and}\qquad y = x^{2} - 4x + 5
    Answer this question yourself

Common questions

Can I expand a product instead of using the product rule?
For polynomials, yes: expanding and differentiating term by term gives the same answer and makes a useful check. It stops being practical once the factors are not polynomials, which is why the rule is worth being fluent in.
Do I need to simplify my derivative?
Enough to make it usable. If the question then asks you to evaluate at a point, find critical points or compare against given options, simplifying first is almost always faster than not.
How do I know which rule to use first?
Ask what the outermost operation is. If the whole expression is one thing divided by another, start with the quotient rule; if it is two things multiplied, start with the product rule; if it is a function applied to an expression, that is the chain rule, which has its own topic.

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