IB Analysis and Approaches HL · Differential calculus · Section 2 of 14

Differentiability and continuity; operations with derivatives

3 worked examples · By Álvaro López

Key idea

Being differentiable at a point is more demanding than being continuous: the graph must not only "not break", it must also be smooth. With this idea you solve the piecewise-function exercises with parameters, which are very common in the IB, and you justify the rules for differentiating sums, products and quotients.

Differentiable implies continuous

The reason: for $x \neq = a$ you can write $f (x) - f (a) = \frac{f ( x ) - f ( a )}{x - a} \cdot (x - a) .$ As $x \to a$ , the first factor tends to $f^{'} (a)$ , which is a number, and the second tends to $0$ . The product tends to $0$ , that is, $f (x) \to f (a)$ .

The converse is false: there are continuous functions that are not differentiable. What is used a lot is the contrapositive: if $f$ is not continuous at $a$ , it is not differentiable at $a$ . There is no need to look at any derivative if there is already a jump.

Points where there is no derivative

The derivative does not exist at a point $a$ when one of these three things happens:

Discontinuity: the graph jumps, or is not defined at $a$ .
Corner point (a "corner"): $f$ is continuous but $f^{'} (a^{-}) \neq = f^{'} (a^{+})$ ; there are two different one-sided tangents. The basic example is $f (x) = ∣ x ∣$ at $x = 0$ : the gradient is $- 1$ on the left and $1$ on the right.
Vertical tangent: $f$ is continuous and the quotient $\frac{f ( a + h ) - f ( a )}{h}$ grows without bound. This happens with $f (x) = 3 x$ at $x = 0$ .

Continuous function with a corner point at (2,1): one-sided tangents with gradients 2 and -1

Piecewise functions

A function with several formulas can be differentiable at the point where it changes formula, but only if it joins up properly and has no corner.

If the function is not continuous at $a$ , step 2 is no longer needed.

Operations with differentiable functions

If $f$ and $g$ are differentiable at a point, so are their sum, their product and their quotient (the latter, if $g \neq = 0$ there).

The sum and constant-multiple rules follow directly from the limit. For the product, it is enough to add and subtract $f (x) g (x + h)$ in the numerator of the definition: $\frac{f ( x + h ) g ( x + h ) - f ( x ) g ( x )}{h} = \frac{f ( x + h ) - f ( x )}{h} g (x + h) + f (x) \frac{g ( x + h ) - g ( x )}{h},$ and as $h \to 0$ we get $f^{'} g + f g^{'}$ (it uses the fact that $g$ , being differentiable, is continuous, so $g (x + h) \to g (x)$ ). The quotient rule follows by applying the product rule to $f \cdot \frac{1}{g}$ .

The derivative of a product is not the product of the derivatives: that is why the rule must be used. In the next section these rules are applied to the usual functions.

Worked examples

Example 1 Basic

A corner in an absolute value

Let $f (x) = ∣ x - 2∣$ .

a) Calculate $f^{'} (2^{-})$ .

b) Calculate $f^{'} (2^{+})$ .

c) Is $f$ differentiable at $x = 2$ ? Answer "yes" or "no".

Solution

Write $f$ piece by piece: $f (x) = 2 - x$ if $x < 2$ , and $f (x) = x - 2$ if $x \geq 2$ . It is continuous at $2$ , because both pieces are $0$ there.

a) On the left $f (x) = 2 - x$ , and its derivative is $- 1$ : $f^{'} (2^{-}) = - 1$ .

b) On the right $f (x) = x - 2$ , and its derivative is $1$ : $f^{'} (2^{+}) = 1$ .

c) The one-sided derivatives do not agree ( $- 1 \neq = 1$ ): the graph has a corner and it is not differentiable at $x = 2$ .

Example 2 Medium

Rules with a table of values

Two differentiable functions $f$ and $g$ have the following data at $x = 2$ :

	$f$	$g$	$f^{'}$	$g^{'}$
$x = 2$	$3$	$- 1$	$4$	$2$

a) Calculate the derivative of $3 f - 2 g$ at $x = 2$ .

b) Calculate the derivative of $f g$ at $x = 2$ .

c) Calculate the derivative of $\frac{f}{g}$ at $x = 2$ .

d) Find the equation of the tangent to the graph of $y = f (x) g (x)$ at $x = 2$ , in the form $y = m x + n$ .

Solution

a) $(3 f - 2 g)^{'} (2) = 3 f^{'} (2) - 2 g^{'} (2) = 3 \cdot 4 - 2 \cdot 2 = 8$ .

b) $(f g)^{'} (2) = f^{'} (2) g (2) + f (2) g^{'} (2) = 4 \cdot (- 1) + 3 \cdot 2 = 2$ .

c) $(\frac{f}{g})^{'} (2) = \frac{f ^{'} ( 2 ) g ( 2 ) - f ( 2 ) g ^{'} ( 2 )}{g ( 2 ) ^{2}} = \frac{4 \cdot ( - 1 ) - 3 \cdot 2}{( - 1 ) ^{2}} = - 10$ .

d) The point is $(2, f (2) g (2)) = (2, - 3)$ and the gradient is $(f g)^{'} (2) = 2$ : $y + 3 = 2 (x - 2) ⟹ y = 2 x - 7.$

Example 3 Hard

Parameters so that a piecewise function is differentiable

Let $f (x) = ⎩ ⎨ ⎧ a x^{2} + b x \frac{1}{x} if x \leq 1, if x > 1,$ with $a, b \in R$ .

a) For $x \to 1^{+}$ , calculate $x \to 1^{+} lim f (x)$ and $f^{'} (1^{+})$ .

b) For $x \to 1^{-}$ , calculate the limit of $f$ and $f^{'} (1^{-})$ in terms of $a$ and $b$ .

c) Find the values of $a$ and $b$ for which $f$ is differentiable at $x = 1$ .

d) With those values, find the equation of the tangent to the graph at $x = 2$ .

Solution

a) For $x > 1$ , $f (x) = \frac{1}{x}$ , so $x \to 1^{+} lim f (x) = 1$ . Its derivative is $- \frac{1}{x ^{2}}$ , and $f^{'} (1^{+}) = - 1$ .

b) For $x \leq 1$ , $f (x) = a x^{2} + b x$ : $x \to 1^{-} lim f (x) = a + b$ . Its derivative is $2 a x + b$ , and $f^{'} (1^{-}) = 2 a + b$ .

c) For it to be differentiable it must be continuous and have equal one-sided derivatives: $a + b 2 a + b = 1 = - 1 (continuity) (one-sided derivatives)$ Subtracting, $a = - 2$ , and then $b = 3$ .

d) At $x = 2$ we are on the piece $\frac{1}{x}$ : $f (2) = \frac{1}{2}$ and $f^{'} (2) = - \frac{1}{4}$ . The tangent is $y - \frac{1}{2} = - \frac{1}{4} (x - 2) ⟹ y = - \frac{x}{4} + 1.$

In the exam: write the two conditions separately (continuity and one-sided derivatives) and say which is which. Many people set up only one of them and are left with an undetermined parameter.

Common mistakes

Summary

Differentiable at $a$ $\Rightarrow$ continuous at $a$ . The converse is false ( $∣ x ∣$ at $0$ ).
There is no derivative at a point if there is a discontinuity, a corner point ( $f^{'} (a^{-}) \neq = f^{'} (a^{+})$ ) or a vertical tangent.
Piecewise function: differentiable at $a$ if it is continuous at $a$ and the one-sided derivatives agree; with parameters, each condition gives an equation.
$(f \pm g)^{'} = f^{'} \pm g^{'}$ , $(k f)^{'} = k f^{'}$ , $(f g)^{'} = f^{'} g + f g^{'}$ , $(\frac{f}{g})^{'} = \frac{f ^{'} g - f g ^{'}}{g ^{2}}$ .

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