Applications of Derivatives notes: ISC Class 12 Mathematics

1. Rate of change

If $y = f(x)$ , then $\dfrac{dy}{dx}$ is the rate of change of $y$ with respect to $x$ ; $f'(x_0)$ is the rate at $x = x_0$ . When $x$ and $y$ both vary with time, $\dfrac{dy}{dt} = \dfrac{dy}{dx}\cdot\dfrac{dx}{dt}$ .

Method for related rates: write the relation between the quantities (remove any extra variable using the given data), differentiate with respect to $t$ , then substitute the values at the instant asked about. Give units, such as cm $^2$ /s.

Marginal cost and revenue: $C'(x)$ and $R'(x)$ are the approximate cost and revenue of one more item.

A derivative measures how fast one quantity changes with another.
Quantity	Rate	Units
Area $A$ of a ripple	$\dfrac{dA}{dt}$	cm $^2$ per second
Edge $x$ of a cube	$\dfrac{dx}{dt}$	cm per second
Cost $C$ of $x$ items	$\dfrac{dC}{dx}$ (marginal cost)	rupees per item

The same ripple grows faster in area as it gets bigger.
Radius $r$ (cm)	$\dfrac{dA}{dt} = 2\pi r \cdot 4$
$4$	$32\pi$
$8$	$64\pi$
$12$	$96\pi$

Remove the extra variable before differentiating.
Step	Working
Volume in $h$	$r = \dfrac h3 \Rightarrow V = \dfrac{\pi h^3}{27}$
Differentiate	$\dfrac{dV}{dt} = \dfrac{\pi h^2}{9}\dfrac{dh}{dt}$
Substitute $h = 9$	$18\pi = 9\pi\dfrac{dh}{dt}$
Answer	$\dfrac{dh}{dt} = 2$ cm/s

Formulas

Chain rule for rates
$\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}$

Key terms

Rate of change: $\dfrac{dy}{dx}$ , how fast $y$ changes per unit change of $x$ .
Marginal cost: $C'(x)$ , the rate of change of total cost with output.

2. Increasing and decreasing functions

For $f$ continuous on $[a, b]$ and differentiable on $(a, b)$ :

$f'(x) > 0$ on $(a, b)$ : $f$ is strictly increasing on $[a, b]$ .
$f'(x) < 0$ on $(a, b)$ : $f$ is strictly decreasing.
$f'(x) = 0$ throughout: $f$ is constant.

Isolated points with $f' = 0$ (as for $x^3$ at $0$ ) do not stop $f$ being strictly increasing. Method: factorise $f'(x)$ , mark its zeros, test the sign on each interval.

Split the line at the zeros of the derivative, then test each piece.
Interval	Sign of $f'(x) = 3(x + 3)(x - 1)$	Behaviour of $f$
$(-\infty, -3)$	$+$	increasing
$(-3, 1)$	$-$	decreasing
$(1, \infty)$	$+$	increasing

$y = \dfrac{x^3}{3} - 4x$

Rising where the slope is positive, falling where it is negative.

$y = x^3 + 3x^2 - 9x + 1$

The turning points separate the increasing and decreasing pieces.

Formulas

Monotonicity test
$f'(x) > 0 \Rightarrow f \text{ strictly increasing}$

Key terms

Strictly increasing: $x_1 < x_2 \Rightarrow f(x_1) < f(x_2)$ .

3. Tangents, normals and the angle between curves

At $(x_1, y_1)$ on $y = f(x)$ with $m = f'(x_1)$ :

Tangent: $y - y_1 = m(x - x_1)$ .
Normal: $y - y_1 = -\dfrac1m(x - x_1)$ (if $m = 0$ the normal is $x = x_1$ ; if the tangent is vertical the normal is $y = y_1$ ).

For parametric curves use $m = \dfrac{dy/dt}{dx/dt}$ .

Angle between curves at a common point: $\tan\theta = \left|\dfrac{m_1 - m_2}{1 + m_1m_2}\right|$ . The curves touch when $m_1 = m_2$ and are orthogonal when $m_1m_2 = -1$ .

$y = x^2 - 2x + 2$
tangent $y = 2x - 2$
normal $y = -\tfrac12 x + 3$

Tangent slope $m$; normal slope $-\dfrac1m$.

Point, slope, then point-slope form twice.
Step	Working
Point	$x = 2$ , $y = 8 - 6 + 1 = 3$
Slope	$\dfrac{dy}{dx} = 3x^2 - 3 = 9$
Tangent	$y - 3 = 9(x - 2)$ , i.e. $y = 9x - 15$
Normal	$y - 3 = -\dfrac19(x - 2)$ , i.e. $x + 9y - 29 = 0$

Turn the words into an equation in $x$.
Condition on the tangent	Equation to solve
parallel to the $x$ -axis	$\dfrac{dy}{dx} = 0$
parallel to the $y$ -axis	$\dfrac{dx}{dy} = 0$
parallel to a line of slope $m$	$\dfrac{dy}{dx} = m$
perpendicular to a line of slope $m$	$\dfrac{dy}{dx} = -\dfrac1m$

$y = x^2$
$y = 6 - x^2$

The angle between curves is the angle between their tangents at a common point.

$y = x$
$y = \sqrt{8 - x^2}$

Slopes $1$ and $-1$ multiply to $-1$: the curves cut at right angles.

Formulas

Angle between curves
$\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$

Key terms

Normal: The line through the point of contact perpendicular to the tangent.
Orthogonal curves: Curves whose tangents at a common point are perpendicular.

4. Local maxima and minima

A critical point $c$ has $f'(c) = 0$ or $f'(c)$ undefined; local extrema occur only there.

First derivative test: $f'$ changes $+ \to -$ : local maximum; $- \to +$ : local minimum; no change: neither.

Second derivative test (when $f'(c) = 0$ ): $f''(c) < 0$ maximum, $f''(c) > 0$ minimum, $f''(c) = 0$ the test fails, so use the first test.

A local maximum value can be smaller than a local minimum value elsewhere (as for $x + \dfrac1x$ ).

Look at the sign of the derivative just before and just after $c$.
Sign of $f'$ as $x$ passes $c$	Conclusion at $c$
$+$ to $-$	local maximum
$-$ to $+$	local minimum
no change	neither (for example a point of inflection)

$y = -x^3 + 3x^2 + 9x - 4$

A local maximum is the top of a hill, not necessarily the highest point overall.

Computed by sampling the derivative on each interval.
Interval	$f'(x) = -3(x + 1)(x - 3)$	Conclusion
$(-\infty, -1)$	$-$
$(-1, 3)$	$+$	local minimum at $x = -1$ , value $-9$
$(3, \infty)$	$-$	local maximum at $x = 3$ , value $23$

$y = x^4$
$y = x^3$

Same zero derivatives at $0$, different behaviour: the second derivative test cannot decide.

Negative second derivative: maximum. Positive: minimum.
Stationary point	$f''(x) = 12x^2 - 8$	Conclusion
$x = 0$	$-8 < 0$	local maximum $f(0) = 1$
$x = \pm\sqrt2$	$16 > 0$	local minimum $f(\pm\sqrt2) = -3$

Formulas

Second derivative test
$f'(c) = 0,\ f''(c) < 0 \Rightarrow \text{local maximum}$

Key terms

Critical point: A point where $f'(c) = 0$ or $f'(c)$ does not exist.
Point of inflection: A point where the curve changes the direction in which it bends.

5. Absolute maxima and minima on a closed interval

A function continuous on $[a, b]$ has an absolute maximum and an absolute minimum there.

Closed-interval method: find the critical points in $(a, b)$ ; evaluate $f$ at each of them and at $a$ and $b$ ; the largest value is the absolute maximum and the smallest the absolute minimum. A value may be attained at more than one point.

$y = x^3 - 3x^2 + 1$ , $-1 \le x \le 3$

Absolute extremes can sit at end points as well as at turning points.

Compare every candidate: largest is the absolute maximum, smallest the absolute minimum.
Candidate $x$	$-1$ (end)	$0$ (critical)	$2$ (critical)	$3$ (end)
$f(x)$	$-3$	$1$	$-3$	$1$

Formulas

Closed-interval method
$\max_{[a,b]} f = \max\{f(a), f(b), f(c_1), \dots\}$

Key terms

Absolute maximum: The greatest value of $f$ on the whole interval.

6. Optimisation problems

Name the quantity $Q$ to be made greatest or least; draw a figure.
Use the given condition to write $Q$ in one variable, and state its domain.
Solve $Q'(x) = 0$ .
Confirm with $Q''$ or a sign change (or by comparing end values).
Answer exactly what is asked, with units.

Useful results: for a fixed perimeter, the rectangle of greatest area is a square; a closed cylinder of fixed volume has least surface area when its height equals its diameter.

Five steps for every word problem.

The side along the wall is twice each fenced end.
Step	Working
Variables	sides $w$ , $w$ fenced; side $80 - 2w$ parallel to the wall
Area	$A(w) = w(80 - 2w) = 80w - 2w^2$ , $0 < w < 40$
Stationary	$A'(w) = 80 - 4w = 0 \Rightarrow w = 20$ ; $A'' = -4 < 0$
Answer	$20$ m by $40$ m, area $800$ m $^2$

$A = 80w - 2w^2$

The top of the hill confirms the maximum found by calculus.

Margins reduce each dimension before you multiply.
Step	Working
Page	width $x$ cm, height $\dfrac{600}{x}$ cm
Printed area	$A(x) = (x - 4)\left(\dfrac{600}{x} - 6\right) = 624 - 6x - \dfrac{2400}{x}$
Stationary	$A'(x) = -6 + \dfrac{2400}{x^2} = 0 \Rightarrow x = 20$ ; $A'' < 0$
Answer	page $20$ cm by $30$ cm; printed area $16 \times 24 = 384$ cm $^2$

Formulas

Optimum condition
$Q'(x) = 0,\ Q''(x) < 0 \Rightarrow \text{maximum}$

Key terms

Optimisation: Finding the greatest or least value of a quantity under given conditions.

Exam tips

How ISC marks are awarded

Related rates (2–3 marks): the relation, its derivative with respect to time, substitution with units.
Intervals (3 marks): factorised $f'(x)$ , sign chart, intervals named.
Tangent and normal (2–3 marks): point, slope, equations.
Local and absolute extrema (3 marks): stationary points, a test with its reason, the values; include end points on a closed interval.
Optimisation (5 marks, in parts): function in one variable, derivative, solution, confirmation, final answer.

Common mistakes

Substituting values before differentiating in related rates.
Forgetting the minus sign in the slope of the normal.
Concluding "no extremum" when $f''(c) = 0$ : the test has failed, not the extremum.
Leaving out the end points when finding absolute extrema.
Answering with $x$ when the question asks for the area or the cost.

1. Rate of change

Formulas

Key terms

2. Increasing and decreasing functions

Formulas

Key terms

3. Tangents, normals and the angle between curves

Formulas

Key terms

4. Local maxima and minima

Formulas

Key terms

5. Absolute maxima and minima on a closed interval

Formulas

Key terms

6. Optimisation problems

Formulas

Key terms

Exam tips

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