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Derivatives and their applications

The derivative measures how fast a function changes: it is the slope of the tangent line to its graph and the instantaneous rate of change. This topic explains its definition, the rules to compute it (product, quotient, chain) and its uses: tangent lines, increase and decrease, maxima and minima and optimization.

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Definition

The slope of the line through $(x, f (x))$ and $(x + h, f (x + h))$ is the difference quotient $\frac{f ( x + h ) - f ( x )}{h}$ . As the two points get closer ( $h \to 0$ ), the secant becomes the tangent:

f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}

Secants of

y = x^{2}

through

P = (1, 1)

: as the second point gets closer they tend to the tangent, of slope

f^{'} (1) = 2

For example, for $f (x) = x^{2}$ : $\frac{( x + h ) ^{2} - x ^{2}}{h} = \frac{2 x h + h ^{2}}{h} = 2 x + h \to 2 x$ . If $x (t)$ is a position, $x^{'} (t)$ is the velocity; if $V$ is a volume, $d V / d t$ is the flow rate. A function with a corner (like $∣ x ∣$ at 0) or a jump is not differentiable there.

Differentiation rules

Function	Derivative	Function	Derivative
$x^{n}$	$n x^{n - 1}$	$sin x$	$cos x$
$e^{x}$	$e^{x}$	$cos x$	$- sin x$
$a^{x}$	$a^{x} ln a$	$tan x$	$sec^{2} x$
$ln x$	$\frac{1}{x}$	$arctan x$	$\frac{1}{1 + x ^{2}}$
$x$	$\frac{1}{2 x}$	$arcsin x$	$\frac{1}{1 - x ^{2}}$

The derivative of a sum is the sum of the derivatives and constants come out. For products, quotients and composite functions:

(uv)^{'} = u^{'} v + u v^{'} (\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}} (f (g (x)))^{'} = f^{'} (g (x)) \cdot g^{'} (x)

The chain rule reads «derivative of the outer function, with the inside untouched, times the derivative of the inside». It is the most forgotten one: $(sin x^{2})^{'} = cos x^{2} \cdot 2 x$ , not $cos x^{2}$ .

Applications

Tangent line at $x = a$ : $y = f (a) + f^{'} (a) (x - a)$ . Near $a$ it is the best linear approximation of the function.

Increase and decrease. Where $f^{'} > 0$ the function increases; where $f^{'} < 0$ , it decreases. Critical points ( $f^{'} = 0$ or $f^{'}$ does not exist) are the candidates for a local maximum or minimum. They are classified by the sign change of $f^{'}$ or by the second derivative: $f^{''} < 0$ is a maximum (curve bending down), $f^{''} > 0$ is a minimum; if $f^{''} = 0$ the test does not decide.

Optimization. Write the quantity to maximize or minimize as a function of one variable, set its physically meaningful interval, find the critical points inside and compare them with the ends of the interval.

The calculator differentiates by the same rules and explains each one («it is a product: product rule…», «outer function, inner function…»). With long expressions it names the parts and differentiates each separately, as on the board.

Common mistakes

Differentiating a product as the product of the derivatives: $(x e^{x})^{'} \neq = 1 \cdot e^{x}$ .
Forgetting the derivative of the inside in the chain rule.
Swapping the order in the numerator of the quotient rule ( $u^{'} v - u v^{'}$ , not $u v^{'} - u^{'} v$ ).
In optimization, keeping a critical point outside the valid interval or not comparing with the ends.

Worked examples

Example 1Product rule

Differentiate

f (x) = x e^{x}

Solution

It is a product with $u = x$ and $v = e^{x}$ : $u^{'} = 1$ , $v^{'} = e^{x}$ .
$f^{'} (x) = u^{'} v + u v^{'} = 1 \cdot e^{x} + x e^{x} = x e^{x} + e^{x} = (x + 1) e^{x}$ .

f^{'} (x) = x e^{x} + e^{x}

Solve it in Simuveliq

In Math → Symbolic calculator, topic Calculus, operation Differentiate: type x e^x and press Compute. The Answer is $f^{'} (x) = x e^{x} + e^{x}$ .

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Example 2Quotient rule

Differentiate

f (x) = \frac{x + 1}{x - 1}

Solution

$u = x + 1$ , $v = x - 1$ , $u^{'} = v^{'} = 1$ .
$f^{'} (x) = \frac{1 \cdot ( x - 1 ) - ( x + 1 ) \cdot 1}{( x - 1 ) ^{2}} = \frac{- 2}{( x - 1 ) ^{2}}$ . It is always negative: the function decreases on each side of $x = 1$ .

f^{'} (x) = - \frac{2}{( x - 1 ) ^{2}}

Solve it in Simuveliq

In Math → Symbolic calculator, topic Calculus, operation Differentiate: type (x+1)/(x-1) and press Compute. The Answer is $- \frac{2}{( x - 1 ) ^{2}}$ .

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Example 3Chain rule

Differentiate

f (x) = sin (x^{2})

Solution

Outer function: sine; inner: $x^{2}$ .
Derivative of the outer one with the inside untouched: $cos (x^{2})$ . Times the derivative of the inside: $2 x$ .
$f^{'} (x) = 2 x cos (x^{2})$ .

f^{'} (x) = 2 x cos (x^{2})

Solve it in Simuveliq

In Math → Symbolic calculator, topic Calculus, operation Differentiate: type sin(x^2) and press Compute. The Answer is $2 x cos (x^{2})$ .
The small graph draws $f$ and $f^{'}$ together: where $f$ has a maximum or a minimum, $f^{'}$ crosses the axis.

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Example 4Tangent line

Find the tangent line to

f (x) = x^{3} - 2 x

x = 1

Solution

$f (1) = 1 - 2 = - 1$ . $f^{'} (x) = 3 x^{2} - 2$ , so $f^{'} (1) = 1$ .
$y = f (1) + f^{'} (1) (x - 1) = - 1 + (x - 1) = x - 2$ .

y = x - 2

Solve it in Simuveliq

In the Compute operation type tangente(x^3-2x, 1) and press Compute: the Answer is $y = x - 2$ , with $f (1) = - 1$ and $f^{'} (1) = 1$ in the steps. The command is in the List of commands.

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Example 5Maxima and minima

Find the local maxima and minima of

f (x) = x^{3} - 3 x

and where it increases or decreases.

Solution

$f^{'} (x) = 3 x^{2} - 3 = 3 (x - 1) (x + 1) = 0$ → critical points $x = - 1$ and $x = 1$ .
$f^{''} (x) = 6 x$ : $f^{''} (- 1) = - 6 < 0$ → maximum, $f (- 1) = - 1 + 3 = 2$ ; $f^{''} (1) = 6 > 0$ → minimum, $f (1) = - 2$ .
Sign of $f^{'}$ : positive outside $[- 1, 1]$ (increasing) and negative inside (decreasing).

Local maximum

(- 1, 2)

and local minimum

(1, - 2)

Solve it in Simuveliq

In Math → Symbolic calculator, topic Calculus, operation Maxima and minima: type x^3-3x and press Compute. The Answer is «max (−1, 2), min (1, −2)».

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Example 6Optimization: the box of maximum volume

Equal squares of side

x

are cut from the corners of a 12 cm square sheet and the edges are folded up to make an open box. Which

x

gives the largest volume?

Solution

The base is $12 - 2 x$ per side and the height is $x$ : $V (x) = x (12 - 2 x)^{2}$ , with $0 < x < 6$ (otherwise there is no box).
$V^{'} (x) = (12 - 2 x)^{2} - 4 x (12 - 2 x) = (12 - 2 x) (12 - 6 x) = 12 x^{2} - 96 x + 144 = 0$ → $x = 2$ or $x = 6$ .
$x = 6$ is the end of the interval (zero volume). At $x = 2$ : $V^{''} (2) = 24 \cdot 2 - 96 = - 48 < 0$ → maximum, $V (2) = 2 \cdot 8^{2} = 128$ cm³. At the ends $V = 0$ , so it is the absolute maximum.

x = 2

cm; maximum volume

128

cm³ (

8 \times 8 \times 2

box).

Solve it in Simuveliq

In Math → Symbolic calculator, topic Calculus, operation Maxima and minima: type x(12-2x)^2 and press Compute. The Answer is «max (2, 128), min (6, 0)».
The calculator studies the function on the whole real line: the minimum $(6, 0)$ is the edge of the physical interval, where the box vanishes.

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Derivatives and their applications

Definition

Differentiation rules

Applications

Common mistakes

Worked examples

Example 1Product rule

Solution

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Example 2Quotient rule

Solution

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Example 3Chain rule

Solution

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Example 4Tangent line

Solution

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Example 5Maxima and minima

Solution

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Example 6Optimization: the box of maximum volume

Solution

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