Applications of Differentiation

Tangents and Normals to Curves #

Gradient for the tangent, flipped for the normal.

The derivative at a point gives the tangent gradient, and the normal is its negative reciprocal

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At x = 2 on $y=x^2$y = x², the derivative 2x gives a tangent gradient of 4.

Put the point (2, 4) and gradient 4 into $y-y_1=m\left(x-x_1\right)$y − y₁ = m(x − x₁): the tangent is y = 4x − 4.

The normal crosses the tangent at a right angle, so its gradient is the negative reciprocal, $-\frac{1}{4}$−1/4.

The Tangent-Line Approximation #

Near the point of contact, use the tangent.

Near the point of contact a curve and its tangent agree closely enough to compute with

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Close to the point of contact, a curve and its tangent are nearly the same line.

That tangent, written as a formula, is L(x) = f(a) + f'(a)(x − a).

For $\sqrt{4.1}$√4.1 take a = 4: L = 2 + 0.25 × 0.1 = 2.025. The true value is 2.0248.

This curve bends down, so its tangent lies above it and L comes out too big.

When the curve bends up, the tangent lies below it instead, so L comes out too small.

So the sign of the second derivative says whether the estimate is too big or too small.

The Mean Value Theorem #

Some tangent runs parallel to the chord.

On a smooth arc some tangent runs parallel to the chord joining the two ends

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Join the ends of the arc: that chord’s gradient is the average rate of change from a to b.

Somewhere strictly between a and b, one tangent runs exactly parallel to that chord.

That is the theorem: f'(c) equals the average rate, for at least one c inside.

The theorem has two conditions: continuous from a to b, and differentiable everywhere between.

Equal end heights make the chord flat, so a flat tangent is promised: Rolle’s theorem.

Drop differentiability and it fails: |x| has a flat chord and no flat tangent at all.

Increasing and Decreasing Functions #

The derivative’s sign says which way it heads.

The sign of the derivative says whether a function is increasing or decreasing

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Left of the minimum the function is decreasing, and the gradient is negative.

Right of the minimum it is increasing, and the gradient is positive.

So the sign of the derivative, not its size, is what gives the direction.

Finding Stationary Points #

Set the derivative to zero and solve.

Setting the derivative to zero finds every point where the curve levels off

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At the minimum itself the tangent is flat, so the gradient there is 0.

So set the derivative to zero and solve. That is the whole method.

The tangent is flat at a maximum too, so solving f' = 0 finds every stationary point without saying which kind it is.

Critical Points and Local Extrema #

Where f' is zero, or missing altogether.

A critical point is where the derivative is zero or missing, and every local extreme is one

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A local maximum is higher than every point close to it, not necessarily the highest on the whole curve.

In this window the right-hand end climbs higher, so that local maximum is not the global maximum.

A critical point is where f' is zero — or where f' does not exist at all.

This cusp has a minimum where f' does not exist, so solving f' = 0 cannot find it.

Every interior local extreme sits at a critical point, so those are the candidates.

The converse can fail: $y=x^3$y = x³ is flat at 0 and still climbs straight through it.

The Second Derivative Test #

Peak or trough? The second derivative knows.

The second derivative says whether a flat point is a peak or a trough

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A trough curves upwards, so its gradient is growing and the second derivative is positive.

A peak curves downwards, so the second derivative there is negative.

And if the second derivative is zero the test is silent — look at the curve instead.

The First Derivative Test #

Read the sign either side.

The sign of f' on either side classifies a flat point even when the second derivative test is silent

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At the flat bottom of $y=x^4$y = x⁴, f'' is 0 — the second derivative test goes silent.

Across the flat point of $x^3$x³, f' runs +, 0, + — no peak, the climb just pauses.

Climb then fall (+, 0, −) is a peak; fall then climb (−, 0, +) is a trough.

The sign of f' on either side always decides, even where the second derivative test cannot.

Absolute Extrema on a Closed Interval #

Compare every critical point against both ends.

On a closed interval the extremes exist and sit at a critical point or an end

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A curve that is continuous on a closed interval reaches a highest and a lowest value.

An extreme can sit at an end, where no tangent is flat — so solving f' = 0 never finds it.

So list every critical point inside, add both ends, and that is every candidate.

For $x^2-4x+3$x² − 4x + 3 on [0, 3]: f(0) = 3, f(2) = −1 and f(3) = 0.

Compare those values: the largest is 3 at x = 0, the smallest is −1 at x = 2.

Leave an end open and the guarantee is lost: $\frac{1}{x}$1/x grows without bound near the missing end.

Points of Inflection #

The bend changes sides.

Where the bend changes sides, f'' passes through zero and changes sign

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$y=x^3$y = x³ curves down on the left of 0 and up on the right — the bend flips.

Where the bend flips, f'' passes through 0 — that point is an inflection.

But f'' = 0 alone is not enough: $x^4$x⁴ has it at 0 and never changes bend.

So check the sign of f'' either side — a flip confirms it, as with f'.

Name the two sides: f'' < 0 is concave down, f'' > 0 is concave up.

Curve Sketching with Derivatives #

Intercepts, stationary points and end behavior.

Intercepts, stationary points and end behavior are enough to sketch a curve

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Start with the intercepts: set y to 0, then set x to 0.

Then find the flat points, and let the second derivative tell the peak from the trough.

Finally ask what happens far to the left and far to the right, and join the pieces up.

Optimization Problems #

One function for the goal, then differentiate.

Write the quantity as one function, then differentiate to find its best value

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A fence of 40 meters has half-perimeter 20, so it makes a rectangle x by 20 − x, whatever x you choose.

Its area is x(20 − x), and that curve has one highest point.

Differentiate: $\frac{dA}{dx}=20-2x$dA/dx = 20 − 2x, which is zero at x = 10 — the rectangle of largest area is a square.

Motion in a Straight Line #

Differentiating displacement gives velocity.

Differentiating displacement gives velocity, and differentiating again gives acceleration

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On a graph of displacement against time, the gradient is the velocity.

Differentiate once more and you have acceleration. Momentarily at rest means v = 0.

Velocity carries a sign for direction. Speed is its size: speed = |v|, never negative.

Related Rates #

One rate rides on another through the chain rule.

Two linked quantities have linked rates: differentiate the link, then multiply the rates

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A balloon inflates: the radius grows, and the volume grows because the radius does.

The chain rule links the rates: $\frac{dV}{dt}$dV/dt equals $\frac{dV}{dr}$dV/dr times $\frac{dr}{dt}$dr/dt — the dr cancels.

A cube of side r has $V=r^3$V = r³. At r = 4, with r growing 2 per second, $\frac{dV}{dt}=48\times 2=96$dV/dt = 48 × 2 = 96.

Every problem of this kind has the same two steps: differentiate the link, then multiply the rates.

Parametric Equations #

A curve told by a clock, not a formula.

A parametric curve gives x and y in terms of t, and each value of t marks one point of the path

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$x=t^2$x = t², y = 2t: each value of t marks one point, and the points trace this curve.

Substitute t = 0, 1, 2, 3 and read off the points — a table is the place to start.

Eliminate t and the Cartesian equation appears: $t=\frac{y}{2}$t = y/2, so $x=\frac{y^2}{4}$x = y²/4.

Parametric Differentiation #

Divide the two time-derivatives.

The gradient of a parametric curve is $\frac{dy}{dt}$dy/dt divided by $\frac{dx}{dt}$dx/dt

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The gradient still means $\frac{dy}{dx}$dy/dx — but here x and y are both functions of t.

By the chain rule $\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}$dy/dx = dy/dt × dt/dx, and $\frac{dt}{dx}$dt/dx is 1 over $\frac{dx}{dt}$dx/dt, so $\frac{dy}{dx}$dy/dx is $\frac{dy}{dt}$dy/dt divided by $\frac{dx}{dt}$dx/dt.

For $x=t^2$x = t², y = 2t: $\frac{dy}{dt}=2$dy/dt = 2 and $\frac{dx}{dt}=2t$dx/dt = 2t, so $\frac{dy}{dx}=2/2t=\frac{1}{t}$dy/dx = 2 / 2t = 1/t.

At t = 2 that gradient is $\frac{1}{2}$1/2 — and the tangent drawn at (4, 4) agrees.

Maclaurin Series #

A function rebuilt from its derivatives at zero.

A function is rebuilt near zero from its derivatives at zero, one term per derivative

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Near 0, $e^x$eˣ is almost 1 + x — the tangent line stays close to the curve there.

Each extra derivative adds one more term: $1+x+\frac{x^2}{2}$1 + x + x²/2 follows the curve further.

The general term: term n is the nth derivative at 0, times $x^n$xⁿ, divided by n factorial.

Every derivative of $e^x$eˣ at 0 equals 1, so $e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots$eˣ = 1 + x + x²/2 + x³/6 + …

$\sin x=x-\frac{x^3}{6}+\dots$sin x = x − x³/6 + … — its first term is the small-angle approximation itself.

Where a Series Is Valid #

Each series holds on its own range of x.

Each standard series holds only on its own range of x, and whatever replaces x must stay inside it

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The formula list gives each series with its range. $e^x$eˣ, sin x and cos x hold for every x; the other two do not.

Whatever replaces x must stay inside the range, so for ln(1 + 3x), solve $-1<3x\le 1$−1 < 3x ≤ 1.

Make the bracket start with 1 first: $\ln \left(2+x\right)=\ln 2+\ln \left(1+\frac{x}{2}\right)$ln(2 + x) = ln 2 + ln(1 + x/2), so $-2<x\le 2$−2 < x ≤ 2.

A sum or product of two series holds only where both do, so take the narrower range: $-\frac{1}{2}<x<\frac{1}{2}$−1/2 < x < 1/2.

L'Hôpital's Rule #

$A\frac{0}{0}$A 0/0 limit settled by two derivatives.

$A\frac{0}{0}$A 0/0 limit equals the limit of derivative over derivative

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$\sin x/x$sin x / x gives 0 over 0 at x = 0 — the squeeze theorem showed that the limit is 1.

Zoom in near a: each part of the fraction is close to its own tangent line through zero.

Divide one tangent by the other and (x − a) cancels: the limit is the ratio of the derivatives.

$\sin x/x$sin x / x becomes $\cos x/1$cos x / 1, and at x = 0 that is 1 — the same answer the squeeze gave.

At x = 1, $\left(x^2-1\right)/\left(x-1\right)$(x² − 1) / (x − 1) differentiates to $2x/1$2x / 1, which is 2 — factoring agrees.

The rule applies only to $\frac{0}{0}$0/0 or $\infty$∞/$\infty$∞. If the denominator is not 0, just substitute.

From the Derivative Back to the Function #

Reading the shape of f from a graph of its gradient.

The sign of f' says where f rises and falls; its zeros and its turning point mark where f turns and bends

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This is the graph of f', not of f: its height at each x is the gradient of f there.

f' is zero at x = −1 and x = 2, so f is flat there: those are its stationary points.

Where f' is above the axis, f is increasing; where f' is below it, f is decreasing.

f' changes from positive to negative at x = −1, so f has a maximum there; from negative to positive at x = 2, a minimum.

The lowest point of f', at $x=\frac{1}{2}$x = 1/2, is where f changes from concave down to concave up: a point of inflection.

Drawn over f', the graph of f rises to a maximum at x = −1, falls to a minimum at x = 2, then rises again.