Application of Derivatives

Five functions. At $x=0$, each has $f^{\prime }(0)=0$, except the last, where the derivative does not exist.

What is happening at the origin in each case?

Read the last three rows again. Two functions share $f^{\prime }(0)=f^{\prime \prime }(0)=0$ and end up with opposite verdicts. A fifth has a perfectly good minimum at a point where no derivative exists at all.

$f^{\prime }(c)=0\text{ is neither necessary nor sufficient for an extremum}$

Only one thing separates them, and it is the same thing in every case: does $f^{\prime }$ change sign there?

$x^3:f^{\prime }=3x^2\ge 0\text{ on both sides}\Rightarrow \text{no sign change}\Rightarrow \text{no extremum}$

$x^4:f^{\prime }=4x^3\text{ goes }-\text{ to }+\Rightarrow \text{sign change}\Rightarrow \text{minimum}$

$|x|:f^{\prime }\text{ goes }-1\text{ to }+1\Rightarrow \text{sign change}\Rightarrow \text{minimum}$

That is the whole unit in one idea.

Ask which of the three the question wants, and the method follows with no decision to make.

1. The Derivative as a Rate of Change

$dy/dx$ is the rate at which $y$ changes per unit change in $x$, at a particular point rather than averaged over an interval.

Positive means the quantities move together; negative means one rises as the other falls.

Approximate change and error

Over a small interval the graph is nearly straight.

$\Delta y\approx f^{\prime }(x)\Delta x$

That turns a derivative into an error estimate, which is how measurement error propagates through a calculation.

If a cube's side is $10$ cm with a possible error of $0.02$ cm, then $V=a^3$ gives $\Delta V\approx 3a^2\Delta a=3(100)(0.02)=6$ cm${}^3$.

As percentages, the side is uncertain by $0.2$ per cent and the volume by $0.6$ per cent, three times as much.

$\frac{\Delta V}{V}=n\frac{\Delta a}{a}\text{for }V=a^n$

A percentage error is multiplied by the power. Cubing triples it, square-rooting halves it.

Illustration 1

Estimate $\sqrt{26}$ without a calculator.

Choose the nearest point where the function is exactly known, which is $25$, and treat the extra $1$ as a small step.

$f(x)=\sqrt{x},f^{\prime }(x)=\frac{1}{2\sqrt{x}},f^{\prime }(25)=\frac{1}{10}$

$\sqrt{26}\approx f(25)+f^{\prime }(25)(1)=5+0.1=5.1$

The true value is $5.09902$, so the estimate is high by $0.00098$, an error of about $0.02$ per cent.

The error is one-sided and predictable: $\sqrt{x}$ is concave down, so its tangent line lies above the curve and every such estimate overshoots. Reading the concavity tells you the direction of the error before you check.

2. Related Rates

Most rate questions give one rate and ask for another. The link is always the chain rule, through the variable both depend on, usually time.

$\frac{dV}{dt}=\frac{dV}{dr}\cdot \frac{dr}{dt}$

The method is fixed, and almost all lost marks here are procedural.

1. Write the geometric relation, using no numerical values yet.
2. Differentiate both sides with respect to $t$, treating every variable as a function of time.
3. Substitute the instantaneous values last.

Trap. Substituting a changing value before differentiating turns a variable into a constant, so its derivative becomes zero and the relation collapses. Genuinely fixed quantities, such as the length of a rigid ladder, may be substituted at any stage.

Illustration 2

An inverted cone of height $10$ m and top radius $5$ m is filled at $2$ cubic metres per minute. How fast is the level rising when the depth is $4$ m?

Write the relation with both variables kept symbolic, and eliminate one using similar triangles.

$\frac{r}{h}=\frac{5}{10}\Rightarrow r=\frac{h}{2}\Rightarrow V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi \frac{h^2}{4}h=\frac{\pi h^3}{12}$

Now differentiate with respect to time, and only then substitute.

$\frac{dV}{dt}=\frac{\pi h^2}{4}\cdot \frac{dh}{dt}\Rightarrow 2=\frac{\pi (16)}{4}\frac{dh}{dt}=4\pi \frac{dh}{dt}$

$\frac{dh}{dt}=\frac{1}{2\pi }\approx 0.159\text{ m per minute}$

Note what substituting $h=4$ too early would have done: $V=\pi (64)/12$ is a constant, its derivative is zero, and the equation becomes $2=0$.

At $h=8$ the same inflow raises the level only $1/8\pi \approx 0.040$ m per minute, four times slower, because the surface area has quadrupled.

Illustration 3

A person $1.8$ m tall walks away from a $6$ m lamp post at $1.5$ m per second. How fast does the tip of their shadow move?

Let $x$ be the distance from the post and $s$ the shadow's length. Similar triangles relate them.

$\frac{6}{x+s}=\frac{1.8}{s}\Rightarrow 6s=1.8x+1.8s\Rightarrow s=\frac{3}{7}x$

The tip of the shadow sits at $x+s$, so differentiate that.

$x+s=\frac{10}{7}x\Rightarrow \frac{d}{dt}(x+s)=\frac{10}{7}(1.5)=\frac{15}{7}\approx 2.14\text{ m per second}$

The answer contains no $x$ at all: the tip moves at a constant speed wherever the person is. The shadow itself lengthens at $\frac{3}{7}(1.5)=0.64$ m per second, and the difference between the two rates is exactly the walking speed.

Trap. "How fast is the shadow lengthening" and "how fast is the tip moving" are different questions with different answers. Decide which length the question is differentiating before writing anything.

3. Increasing and Decreasing Functions

$f^{\prime }(x)>0\text{ on an interval}\Rightarrow f\text{ strictly increasing there}$

$f^{\prime }(x)<0\text{ on an interval}\Rightarrow f\text{ strictly decreasing there}$

Trap. Monotonicity is a property of an interval, not of a point. Saying a function is increasing at $x=2$ is meaningless in this syllabus.

A derivative may vanish at isolated points without breaking strict monotonicity. What matters is that it does not stay zero over an interval.

Illustration 4

Show that $f(x)=x-\sin x$ is strictly increasing on the whole real line.

$f^{\prime }(x)=1-\cos x$

Since $\cos x\le 1$ always, $f^{\prime }(x)\ge 0$ everywhere, so the function never decreases.

It equals zero only where $\cos x=1$, that is at $x=0,\pm 2\pi ,\pm 4\pi ,\dots$, which are isolated points, not an interval.

$f^{\prime }\ge 0\text{ with zeros only at isolated points}\Rightarrow \text{strictly increasing}$

The graph confirms it: at each multiple of $2\pi$ the curve flattens momentarily and then carries on upwards, exactly as $x^3$ does at the origin. A function stops being strictly increasing only if $f^{\prime }$ is zero across a whole interval, which would make it flat there.

Illustration 5

Find the intervals of increase and decrease of $f(x)=\frac{\ln x}{x}$, and deduce which is larger, $e^{\pi }$ or ${\pi }^e$.

The domain is $x>0$. Differentiate by the quotient rule.

$f^{\prime }(x)=\frac{\frac{1}{x}\cdot x-\ln x}{x^2}=\frac{1-\ln x}{x^2}$

The denominator is positive throughout the domain, so the sign is entirely the numerator's.

$1-\ln x>0⟺x<e$

So $f$ increases on $(0,e)$ and decreases on $(e,\infty )$, with a maximum at $x=e$ of value $1/e$.

Now use it. Since $e<\pi$ and $f$ decreases beyond $e$:

$f(e)>f(\pi )\Rightarrow \frac{\ln e}{e}>\frac{\ln \pi }{\pi }\Rightarrow \pi \ln e>e\ln \pi \Rightarrow \ln e^{\pi }>\ln {\pi }^e$

$e^{\pi }>{\pi }^e$

The numbers are $23.14$ and $22.46$, so the margin is genuinely small, and no amount of estimating would have settled it. Monotonicity did.

4. Critical Points and the Two Tests

Critical point. A point of the domain where $f^{\prime }(x)=0$ or $f^{\prime }(x)$ fails to exist.

The second half of that definition is what catches $|x|$, and it is routinely forgotten.

The first derivative test

Examine the sign of $f^{\prime }$ on either side of the critical point.

The second derivative test

$f^{\prime }(c)=0\text{ and }{f}^{\prime \prime }(c)<0\Rightarrow \text{maximum};f^{\prime }(c)=0\text{ and }{f}^{\prime \prime }(c)>0\Rightarrow \text{minimum}$

It is faster when it works, but $f^{\prime \prime }(c)=0$ is inconclusive and the first derivative test must then be used.

Concavity and inflection

$f^{\prime \prime }>0$ means concave up; $f^{\prime \prime }<0$ means concave down. An inflection is where the concavity changes, so $f^{\prime \prime }=0$ is necessary but not sufficient: the second derivative must actually change sign.

$x^4:f^{\prime \prime }=12x^2\ge 0\text{ always}\Rightarrow f^{\prime \prime }(0)=0\text{ but no inflection}$

Illustration 6

Find the inflection points of $f(x)=x^4-6x^2$, and say why $x^4$ alone has none.

$f^{\prime }(x)=4x^3-12x,f^{\prime \prime }(x)=12x^2-12=12(x-1)(x+1)$

$f^{\prime \prime }=0\Rightarrow x=\pm 1$

Now test whether the sign actually changes, which is the part that matters. The factor $12(x-1)(x+1)$ is positive outside $[-1,1]$ and negative inside.

$x<-1:f^{\prime \prime }>0;-1<x<1:f^{\prime \prime }<0;x>1:f^{\prime \prime }>0$

Both crossings are genuine sign changes, so there are inflections at $x=\pm 1$, where $f=1-6=-5$.

For $f(x)=x^4$ the second derivative is $12x^2$, which is zero at the origin but positive on both sides. No sign change, so no inflection: the curve is concave up throughout and merely flattens momentarily.

Illustration 7

Examine $f(x)=x^4-4x^3$ for local extrema.

$f^{\prime }(x)=4x^3-12x^2=4x^2(x-3)\Rightarrow \text{critical points }x=0,x=3$

Try the second derivative test first, since it is quicker.

$f^{\prime \prime }(x)=12x^2-24x,f^{\prime \prime }(3)=108-72=36>0\Rightarrow \text{local minimum at }x=3$

$f^{\prime \prime }(0)=0\Rightarrow \text{inconclusive}$

Fall back on the sign of $f^{\prime }$ near $0$. The factor $4x^2$ is positive on both sides, so the sign is decided entirely by $(x-3)$, which is negative on both sides of $0$.

$f^{\prime }<0\text{ for }x<0\text{ and }{f}^{\prime }<0\text{ for }0<x<3\Rightarrow \text{no sign change}$

So $x=0$ is neither a maximum nor a minimum. It is a horizontal point of inflection: the curve flattens and carries on falling, exactly as $x^3$ does at the origin.

The minimum value is $f(3)=81-108=-27$.

5. Absolute Maxima and Minima

On a closed interval, a continuous function attains both an absolute maximum and an absolute minimum, and each occurs either at a critical point or at an endpoint.

$\text{evaluate }f\text{ at every critical point and at both endpoints, then compare}$

Trap. Endpoints are not critical points and the derivative need not vanish there, but they are legitimate places for the absolute extremum. Omitting them is the commonest error in this section.

Illustration 8

Find the absolute maximum and minimum of $f(x)=\sin x+\cos x$ on $[0,\pi ]$.

$f^{\prime }(x)=\cos x-\sin x=0\Rightarrow \tan x=1\Rightarrow x=\frac{\pi }{4}$

Evaluate at that critical point and at both ends.

$\text{absolute maximum }\sqrt{2}\text{ at }x=\pi /4,\text{absolute minimum }-1\text{ at }x=\pi$

The minimum sits at an endpoint, where the derivative is $-1$ rather than $0$. Anyone who checked only the critical point would have reported $\sqrt{2}$ as the maximum and had nothing at all to offer for the minimum.

6. Optimisation Problems

The procedure never varies.

1. Name the quantity to be optimised and write it as a formula.
2. Use the constraint to reduce it to one variable, noting the valid range.
3. Differentiate, set to zero, and solve.
4. Confirm it is the right kind of extremum, and check the endpoints of the range.

Illustration 9

Find the cylinder of greatest volume that can be inscribed in a sphere of radius $R$.

Let the cylinder have radius $r$ and height $h$. The constraint is that its corners touch the sphere, which by Pythagoras in the axial cross-section gives a relation between them.

$r^2+\frac{h^2}{4}=R^2\Rightarrow r^2=R^2-\frac{h^2}{4}$

Substitute to leave one variable, choosing $h$ because $r$ appears only as $r^2$.

$V=\pi r^{2}h=\pi h\left(R^2-\frac{h^2}{4}\right)=\pi \left(R^{2}h-\frac{h^3}{4}\right)$

$\frac{dV}{dh}=\pi \left(R^2-\frac{3h^2}{4}\right)=0\Rightarrow h=\frac{2R}{\sqrt{3}}$

$\frac{d^{2}V}{d{h}^2}=-\frac{3\pi h}{2}<0\Rightarrow \text{maximum}$

$V_{\max }=\pi \cdot \frac{2R}{\sqrt{3}}\left(R^2-\frac{R^2}{3}\right)=\frac{4\pi R^3}{3\sqrt{3}}$

Compare with the sphere's own volume $\frac{4}{3}\pi R^3$: the ratio is $1/\sqrt{3}\approx 0.577$, so the best cylinder fills about $58$ per cent of the sphere.

Illustration 10

A closed cylindrical can must hold a fixed volume $V$. What shape uses the least metal?

Surface area is what to minimise, and the volume is the constraint that removes one variable.

$S=2\pi r^2+2\pi rh,V=\pi r^{2}h\Rightarrow h=\frac{V}{\pi r^2}$

$S=2\pi r^2+\frac{2V}{r}$

$\frac{dS}{dr}=4\pi r-\frac{2V}{r^2}=0\Rightarrow r^3=\frac{V}{2\pi }$

Rather than computing $r$ numerically, substitute back and see what shape emerges.

$h=\frac{V}{\pi r^2}=\frac{2\pi r^3}{\pi r^2}=2r$

$\text{height}=\text{diameter}$

The answer is a proportion rather than a number, so it holds for every volume. That is what makes it worth remembering, and why real cans, which are taller than this, are shaped by printing and handling rather than by metal cost.

7. A Note on Syllabus Emphasis

The current unit text lists exactly three applications: rate of change of quantities, monotonic increasing and decreasing functions, and maxima and minima of functions of one variable.

All three unnamed topics remain examinable in JEE Advanced. The equation of a tangent is still worth the two minutes it takes to learn, since it is a direct reading of the derivative as a slope, but it should not displace practice on optimisation, where the marks in this unit actually sit.

Illustration 11

Beyond JEE Main, for Advanced sitters: prove that $|\sin a-\sin b|\le |a-b|$ for all real $a$ and $b$.

Lagrange's mean value theorem says that for a function differentiable on an interval, some interior point has slope equal to the average slope across it.

$\frac{\sin a-\sin b}{a-b}=\cos c\text{for some }c\text{ between }a\text{ and }b$

Take moduli, and use the fact that a cosine never exceeds $1$ in size.

$\left|\frac{\sin a-\sin b}{a-b}\right|=|\cos c|\le 1\Rightarrow |\sin a-\sin b|\le |a-b|$

The inequality says the sine graph is never steeper than a line of gradient $1$, which is visible in its shape. Turning a visual fact into a proof is what the mean value theorem is for.

Summary

One number, three readings: the value says how fast, the sign says which way, a change of sign says where it turns.

$f^{\prime }(c)=0$ is neither necessary nor sufficient for an extremum. Only a sign change in $f^{\prime }$ decides, which is why $x^3$ and $x^4$ behave differently at the origin and why $|x|$ has a minimum with no derivative at all.

$\Delta y\approx f^{\prime }(x)\Delta x$ estimates values and propagates errors, and a percentage error is multiplied by the power: cubing triples it.

For related rates, write the relation symbolically, differentiate with respect to time, then substitute. Substituting early makes a variable constant and collapses the equation.

Eliminate variables using the geometry before differentiating, as with $r=h/2$ in a cone.

Monotonicity is a property of an interval, not of a point. Isolated zeros of $f^{\prime }$ do not break strict monotonicity, but a whole interval of them does.

A sign analysis of $f^{\prime }$ solves inequality-flavoured questions such as which of $e^{\pi }$ and ${\pi }^e$ is larger.

Critical points include those where $f^{\prime }$ fails to exist, not only where it vanishes.

The second derivative test is faster when it works, and is inconclusive whenever $f^{\prime \prime }(c)=0$; the first derivative test always works.

$f^{\prime \prime }=0$ is necessary but not sufficient for an inflection, since $f^{\prime \prime }$ must actually change sign.

On a closed interval, compare the values at every critical point and at both endpoints; the absolute extremum is frequently at an endpoint.

For optimisation, reduce to one variable using the constraint, note the valid range, then differentiate. Answers that come out as proportions, such as height equalling diameter, hold for every size.