Application of Derivatives

Which point of the parabola $y=x^2$ is closest to $(0,2)$? The vertex, surely — it is the only point the axis of symmetry singles out.

It is not. The two closest points are at $x=\pm \sqrt{3/2}$, a distance $\frac{\sqrt{7}}{2}\approx 1.32$ away, while the vertex is a full $2$ away.

Write $D=x^2+{\left(x^2-2\right)}^2$ and substitute $t=x^2\ge 0$, giving $D=t^2-3t+4$. This is a parabola in $t$ with minimum at $t=\frac{3}{2}$, which is admissible, so $x^2=\frac{3}{2}$ and $D=\frac{7}{4}$.

What makes the question worth asking is that the answer changes character with the height of the point. Replacing $2$ by a general $c>0$ gives $D=t^2+(1-2c)t+c^2$, whose unconstrained minimum sits at $t=c-\frac{1}{2}$. That value is admissible only when $c\ge \frac{1}{2}$; below that threshold the minimum of $D$ over $t\ge 0$ occurs at the boundary $t=0$, and the vertex genuinely is the nearest point.

So the correct answer is a pair of cases, not a formula. Recognising that a constraint can push the optimum to the boundary of its domain is the single most examined idea in this chapter, and it is the reason candidates who differentiate correctly still lose the marks.

1. Tangents, normals and the angle between curves

At a point $(x_1,y_1)$ on a curve, the tangent has slope $m={\frac{dy}{dx}|}_{(x_1,y_1)}$ and the normal has slope $-\frac{1}{m}$. Everything about tangency reduces to those two lines and the conditions imposed on them.

The angle $\theta$ between two curves at a common point is the angle between their tangents there:

$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|,$

so the curves are orthogonal exactly when $m_1{m}_2=-1$, and they touch when $m_1=m_2$. A question asking for the condition that two families cut at right angles is asking for that product to be $-1$ at every common point.

Illustration 1

Show that the curves $xy=c^2$ and $x^2-y^2=a^2$ intersect orthogonally.

Differentiating implicitly, the first gives $y+x\frac{dy}{dx}=0$, so $m_1=-\frac{y}{x}$. The second gives $2x-2y\frac{dy}{dx}=0$, so $m_2=\frac{x}{y}$. Their product is $-1$ at every point, so the two families are orthogonal wherever they meet — no common point ever needs to be found.

Illustration 2

Find the equations of the tangents to $y=x^3$ that pass through $(2,0)$.

The point of contact is unknown, so call it $\left(t,t^3\right)$. The tangent there has slope $3t^2$, and it passes through $(2,0)$ when

$\frac{t^3-0}{t-2}=3t^2\Rightarrow t^3=3t^3-6t^2\Rightarrow 2t^2(t-3)=0.$

So $t=0$ or $t=3$, giving the tangents $y=0$ and $y=27x-54$. Parametrising the point of contact, rather than the line, is what makes external-tangent questions routine.

2. Monotonicity, stated correctly

A differentiable function is increasing on an interval when $f^{\prime }\ge 0$ there, and strictly increasing when in addition $f^{\prime }$ vanishes only at isolated points. That refinement matters: $f(x)=x^3$ has $f^{\prime }(0)=0$ yet is strictly increasing everywhere, because the derivative is positive on both sides and the single zero cannot flatten an interval.

The condition is about an interval, not a point. A function can have $f^{\prime }(a)>0$ at a point without being increasing on any interval around it, which is another consequence of a derivative not having to be continuous.

Illustration 3

For which $a$ is $f(x)=x^3-3a{x}^2+3x+1$ increasing on all of $\mathbb{R}$?

We need $f^{\prime }(x)=3x^2-6ax+3\ge 0$ for every $x$. A quadratic with positive leading coefficient is non-negative everywhere exactly when its discriminant is at most zero:

$36a^2-36\le 0\Rightarrow -1\le a\le 1.$

At the endpoints the derivative touches zero at a single point, which is permitted, so the interval is closed.

3. The three places an extremum can hide

On a closed interval, the greatest and least values of a continuous function occur at one of three kinds of point: an interior point where $f^{\prime }=0$, an interior point where $f^{\prime }$ fails to exist, or an endpoint. Checking only the first is the standard way to lose an optimisation question.

That the greatest and least values exist at all is itself a theorem, and it needs both hypotheses: the function continuous, and the interval closed and bounded. On the open interval $(0,1)$ the function $f(x)=x$ has no greatest value, since it approaches $1$ without reaching it, and on $[0,1]$ a function with a jump can miss its supremum in the same way.

A question set on an open interval or an unbounded one is therefore asking for a supremum rather than a maximum, and the honest answer may well be that no maximum is attained anywhere on the given domain.

To classify an interior critical point, the first derivative test reads the sign of $f^{\prime }$ on either side and is always decisive. The second derivative test is quicker — $f^{\prime \prime }(c)>0$ means a minimum and $f^{\prime \prime }(c)<0$ a maximum — but it is silent when $f^{\prime \prime }(c)=0$, and silence is not evidence.

Illustration 4

Classify the critical point of $f(x)=x^4$ at the origin.

Here $f^{\prime }(0)=0$ and $f^{\prime \prime }(0)=0$, so the second derivative test says nothing. The first derivative test settles it at once: $f^{\prime }(x)=4x^3$ is negative for $x<0$ and positive for $x>0$, so the origin is a minimum. Compare $g(x)=x^3$, where the same two derivatives vanish but the sign of $g^{\prime }$ does not change, so there is no extremum at all.

Illustration 5

Find the greatest and least values of $f(x)=x^3-9x^2+24x$ on $[0,5]$.

$f^{\prime }(x)=3x^2-18x+24=3(x-2)(x-4)$, so the interior critical points are $x=2$ and $x=4$. Evaluating at all four candidates,

$f(0)=0,f(2)=20,f(4)=16,f(5)=20.$

The least value is $0$ at the left endpoint and the greatest is $20$, attained twice — once at an interior maximum and once at an endpoint. Had only the critical points been checked, the least value would have been reported as $16$.

Illustration 6

A right circular cylinder is inscribed in a sphere of radius $R$. Find its greatest volume.

With half-height $h$, the radius satisfies $r^2=R^2-h^2$, so

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

Then $\frac{dV}{dh}=2\pi \left(R^2-3h^2\right)$, vanishing at $h=\frac{R}{\sqrt{3}}$, and $\frac{d^{2}V}{d{h}^2}=-12\pi h<0$ there, confirming a maximum. Substituting gives

$V_{\max }=\frac{4\pi R^3}{3\sqrt{3}}.$

The endpoints $h=0$ and $h=R$ both give zero volume, so the interior critical point is genuinely the answer — but saying so requires having looked.

4. Concavity, inflection and the shape of a curve

The sign of $f^{\prime \prime }$ describes bending: positive means concave upwards, negative concave downwards. A point of inflection is where the concavity changes, which requires $f^{\prime \prime }$ to vanish or fail to exist there and to change sign across it. A vanishing second derivative alone is not enough, exactly as a vanishing first derivative alone does not give an extremum.

Sketching a curve is then a matter of assembling five pieces of information: the domain, the intercepts, the asymptotes, the sign of $f^{\prime }$, and the sign of $f^{\prime \prime }$.

Illustration 7

Locate the points of inflection of $f(x)=x^4-4x^3$.

$f^{\prime \prime }(x)=12x^2-24x=12x(x-2)$, which vanishes at $x=0$ and $x=2$. Since $f^{\prime \prime }$ is a quadratic with distinct roots, it changes sign at each, so both are genuine inflection points. Note that $f^{\prime }(0)=0$ as well, so the origin is a stationary point of inflection — flat, but not an extremum.

Asymptotes complete the picture and are found by three separate limits. A vertical asymptote sits where the function is unbounded, typically at a zero of a denominator that is not cancelled by the numerator. A horizontal asymptote is a finite value of ${\lim }_{x\to \pm \infty }f(x)$. When that limit is infinite, an oblique asymptote $y=mx+c$ may still exist, and the quickest route to it is polynomial division: whatever remains after the linear part tends to zero.

Illustration 8

Sketch $y=\frac{x^2}{x-1}$, identifying its asymptotes and extrema.

Dividing, $y=x+1+\frac{1}{x-1}$, so the oblique asymptote is $y=x+1$ and there is a vertical asymptote at $x=1$. Then

$\frac{dy}{dx}=1-\frac{1}{(x-1{)}^2},$

which vanishes when $(x-1{)}^2=1$, that is at $x=0$ and $x=2$, with values $y=0$ and $y=4$.

The sign of the derivative shows that $x=0$ is a local maximum and $x=2$ a local minimum — and the local maximum value $0$ is smaller than the local minimum value $4$. There is no contradiction: the two lie on different branches, separated by the vertical asymptote, and "local" means only local. A candidate who assumes a maximum must exceed a minimum will misread this curve entirely.

5. Monotonicity as a proof technique

Most inequalities in this chapter are proved by showing that a difference is zero at one endpoint and monotone thereafter. Define $g$ as the difference of the two sides, show $g(0)=0$, and show $g^{\prime }>0$ beyond.

Illustration 9

Prove that $\sin x<x$ for every $x>0$.

Let $g(x)=x-\sin x$. Then $g(0)=0$ and $g^{\prime }(x)=1-\cos x\ge 0$, with equality only at isolated multiples of $2\pi$. So $g$ is strictly increasing on $[0,\infty )$ and therefore positive for $x>0$.

This is where the isolated-zero refinement earns its keep: $g^{\prime }$ does vanish, infinitely often, yet the conclusion is still strict.

Illustration 10

Prove that $e^x\ge 1+x$ for every real $x$, with equality only at $x=0$.

Let $g(x)=e^x-1-x$, so $g(0)=0$ and $g^{\prime }(x)=e^x-1$. The derivative is negative for $x<0$ and positive for $x>0$, so $g$ decreases to a minimum at the origin and increases after. The minimum value is $g(0)=0$, so $g\ge 0$ everywhere with equality only there.

Notice that a single monotonicity argument would not have worked, because $g$ is not monotone; splitting at the critical point was necessary.

Illustration 11

Prove that $\frac{\tan x}{x}>1$ for $0<x<\frac{\pi }{2}$.

Let $g(x)=\tan x-x$. Then $g(0)=0$ and $g^{\prime }(x)={\sec }^{2}x-1={\tan }^{2}x>0$ on the open interval. So $g$ is strictly increasing from zero and hence positive, and dividing by the positive $x$ gives the result.

6. Counting roots from a shape

The number of real roots of $f(x)=0$ is the number of times the graph crosses the axis, and the graph's shape is fixed by its critical values. For a cubic with two critical points, three real roots require the two critical values to have opposite signs.

Illustration 12

For how many real values does $x^3-3x+k=0$ have three distinct real roots?

Let $f(x)=x^3-3x+k$, with $f^{\prime }(x)=3x^2-3$ vanishing at $x=\pm 1$. The local maximum is $f(-1)=2+k$ and the local minimum is $f(1)=k-2$.

Three distinct roots need the maximum above the axis and the minimum below:

$(2+k)>0\text{ and }(k-2)<0\Rightarrow -2<k<2.$

At $k=\pm 2$ one critical value touches the axis and two roots merge; outside the interval only one real root survives. Reading the answer from the two critical values is far quicker than any attempt at the cubic's discriminant.

Illustration 13

Show that $e^x=kx$ has no solution for $0<k<e$, exactly one for $k=e$, and two for $k>e$.

Rearranged, the question asks how often the horizontal line $k$ meets $h(x)=\frac{e^x}{x}$ for $x>0$. Now $h^{\prime }(x)=\frac{e^x(x-1)}{x^2}$, which is negative on $(0,1)$ and positive after, so $h$ falls to a minimum $h(1)=e$ and rises thereafter, tending to infinity at both ends. The line therefore misses the graph below $e$, touches it at $e$, and cuts it twice above.

7. Rates of change and approximation

If two quantities are related, differentiating the relation with respect to time relates their rates. The only skill is writing the relation before differentiating, and eliminating any variable that is not needed.

For small changes, $\Delta y\approx f^{\prime }(x)\Delta x$, which turns a hard evaluation into an easy one plus a correction.

Illustration 14

A spherical balloon is inflated so its volume grows at $20\mathrm{c}{\mathrm{m}}^3/s$. How fast is the radius growing when $r=5$ cm?

From $V=\frac{4}{3}\pi r^3$, differentiating with respect to time gives $\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$, so

$\frac{dr}{dt}=\frac{20}{4\pi (25)}=\frac{1}{5\pi }cm/s.$

The rate falls as the balloon grows, since the same volume must spread over a larger surface — a check that the algebra agrees with the physics.

Illustration 15

Approximate $\sqrt[3]{26}$.

Take $f(x)=x^{1/3}$ at $x=27$, where $f(27)=3$ and $f^{\prime }(27)=\frac{1}{3}(27{)}^{-2/3}=\frac{1}{27}$. With $\Delta x=-1$,

$\sqrt[3]{26}\approx 3-\frac{1}{27}\approx \mathrm{2.9630.}$

The true value is $2.96250$, so the error is about $5\times 10^{-4}$, consistent with a second-order term of size $\frac{1}{2}\left|f^{\prime \prime }\right|\Delta x^2$.

Summary

An optimisation problem is not finished when the derivative is set to zero. The critical point must be admissible, and when a substitution introduces a restriction such as $t=x^2\ge 0$, the optimum can be pushed to the boundary — which is why the nearest point on $y=x^2$ to $(0,c)$ is the vertex for $c\le \frac{1}{2}$ and a symmetric pair above it.

On a closed interval, the greatest and least values live at interior stationary points, at interior points where the derivative does not exist, or at the endpoints, and all three must be evaluated and compared. The second derivative test is quicker than the first but silent when $f^{\prime \prime }$ vanishes, and silence is not evidence: $x^4$ and $x^3$ have the same first two derivatives at the origin and behave completely differently.

Monotonicity requires $f^{\prime }\ge 0$ on an interval, and isolated zeros of $f^{\prime }$ do not spoil strictness. That refinement makes monotonicity the standard way to prove an inequality: set $g$ equal to the difference, check that it vanishes at an endpoint, and show the derivative keeps its sign — splitting at a critical point when the difference is not monotone throughout.

Asymptotes are found by three limits — a vertical one at an uncancelled zero of a denominator, a horizontal one from the limit at infinity, and an oblique one from polynomial division. When a vertical asymptote separates two branches, a local maximum on one can sit below a local minimum on the other, which is a standard way to mislead.

The number of real roots is read from the critical values: a cubic has three distinct roots exactly when its local maximum and local minimum straddle the axis. Tangency and orthogonality are conditions on slopes, and a tangent from an external point is found by parametrising the point of contact rather than the line.