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🇨🇩 Calculus

Derivatives and curve sketching

⏱ 30 minOfficial programme: CD-H3-ANA-02

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🎯 In this lesson, you will learn to

  • Differentiate; use second derivative; full study of a function.
  • Solve optimisation problems.

The lesson

Introduction

In Matadi, a tinsmith cuts squares from the corners of a metal sheet to fold a box: which cut size gives the largest volume? To answer, we must know where a function increases, where it decreases and where it reaches its maximum. The derivative is the tool that gives this information; it also allows the full study of a function and an accurate sketch of its graph.

Explanation

Derivative and rules

The derivative f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a)=\lim\limits_{h\to0}\dfrac{f(a+h)-f(a)}{h} is the slope of the tangent at the point with abscissa aa. The tangent has equation y=f′(a)(x−a)+f(a)y=f'(a)(x-a)+f(a).

FunctionDerivative
xnx^nnxn−1nx^{n-1}
u+vu+v; kukuu′+v′u'+v'; ku′ku'
uvuvu′v+uv′u'v+uv'
uv\dfrac uvu′v−uv′v2\dfrac{u'v-uv'}{v^2}
unu^nnu′un−1nu'u^{n-1}
u\sqrt uu′2u\dfrac{u'}{2\sqrt u}
sin⁡u\sin u; cos⁡u\cos uu′cos⁡uu'\cos u; −u′sin⁡u-u'\sin u

Explanation

Variation, extrema, concavity

On an interval II:

  • if f′(x)>0f'(x)>0 on II (except at isolated points), ff is strictly increasing; if f′(x)<0f'(x)<0, it is strictly decreasing;
  • if f′f' is zero at aa and changes sign, ff has a local extremum at aa (maximum if f′f' goes from ++ to −-, minimum if f′f' goes from −- to ++).

The second derivative f′′f'' gives the concavity: if f′′>0f''>0, the graph is concave up; if f′′<0f''<0, it is concave down. An inflection point is a point where f′′f'' is zero and changes sign: there the graph crosses its tangent.

Explanation

Full study of a function

  1. Domain; parity and periodicity, if any.
  2. Limits at the ends of the domain and asymptotes.
  3. Derivative, sign of f′f', variation table.
  4. Second derivative, concavity, inflection points.
  5. Special points (intersections with the axes), useful tangents, then the sketch.

Optimisation: express the quantity to optimise as a function of a single variable, state the interval of possible values, then find the maximum or minimum with the variation table.

Worked example

Problem: Study the variation of f(x)=x3−3x2−9x+5f(x)=x^3-3x^2-9x+5 and find its inflection point.

Step 1: f′(x)=3x2−6x−9=3(x2−2x−3)=3(x+1)(x−3)f'(x)=3x^2-6x-9=3(x^2-2x-3)=3(x+1)(x-3).

Step 2: f′>0f'>0 on (−∞,−1)(-\infty, -1) and on (3,+∞)(3, +\infty); f′<0f'<0 on (−1,3)(-1, 3).

Step 3: Local maximum f(−1)=−1−3+9+5=10f(-1)=-1-3+9+5=10; local minimum f(3)=27−27−27+5=−22f(3)=27-27-27+5=-22.

Step 4: f′′(x)=6x−6f''(x)=6x-6 is zero at x=1x=1 and changes sign; f(1)=1−3−9+5=−6f(1)=1-3-9+5=-6.

Answer: ff increases, then decreases on [−1,3][-1, 3], then increases; maximum 1010, minimum −22-22; inflection point I(1,−6)I(1, -6).

Worked example

Problem: Find the equation of the tangent to the graph of g(x)=2x+1x−1g(x)=\dfrac{2x+1}{x-1} at the point with abscissa 22.

Step 1: g′(x)=2(x−1)−(2x+1)⋅1(x−1)2=−3(x−1)2g'(x)=\dfrac{2(x-1)-(2x+1)\cdot1}{(x-1)^2}=\dfrac{-3}{(x-1)^2}.

Step 2: g(2)=51=5g(2)=\dfrac{5}{1}=5 and g′(2)=−31=−3g'(2)=\dfrac{-3}{1}=-3.

Step 3: y=g′(2)(x−2)+g(2)=−3(x−2)+5y=g'(2)(x-2)+g(2)=-3(x-2)+5.

Answer: y=−3x+11y=-3x+11.

Worked example

Problem: Bahati has 4040 m of wire mesh to fence a rectangular chicken run against a wall (the wall closes one side). Which dimensions give the largest area?

Step 1: Let xx be the length (in m) of each of the two sides perpendicular to the wall. The side parallel to the wall measures 40−2x40-2x, with 0<x<200<x<20.

Step 2: Area: A(x)=x(40−2x)=40x−2x2A(x)=x(40-2x)=40x-2x^2.

Step 3: A′(x)=40−4xA'(x)=40-4x is zero at x=10x=10; A′>0A'>0 before, A′<0A'<0 after: it is a maximum.

Answer: 1010 m by 2020 m, for a maximum area of 200200 m².

Watch out

Watch out: f′(a)=0f'(a)=0 is not enough for an extremum: for f(x)=x3f(x)=x^3, f′(0)=0f'(0)=0 but f′f' does not change sign, and ff is increasing on R\mathbb{R}. Always study the sign of f′f' around aa. Likewise, f′′(a)=0f''(a)=0 gives an inflection point only if f′′f'' changes sign.

Key points

  • Tangent at aa: y=f′(a)(x−a)+f(a)y=f'(a)(x-a)+f(a).
  • (uv)′=u′v+uv′(uv)'=u'v+uv'; (uv)′=u′v−uv′v2\left(\dfrac uv\right)'=\dfrac{u'v-uv'}{v^2}; (un)′=nu′un−1(u^n)'=nu'u^{n-1}.
  • Sign of f′f': variation; sign change of f′f': extremum.
  • Sign of f′′f'': concavity; sign change of f′′f'': inflection point.
  • Optimisation: one variable, one interval, one variation table.

✍️ Exercises

Answer one exercise at a time. Your answer is checked right away, with an explanation.

Exercise 1 of 8

★ Easy

What is the derivative of f(x)=4x3−5x+2f(x)=4x^3-5x+2?

Your answer