Rules of Differentiation

The Derivative As a Limit #

Where the chord gradient settles.

The derivative is defined as a limit, not as an approximation to one

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A chord through two points has an ordinary rise-over-run gradient.

Slide one end closer to the other and the chord swings toward the tangent.

The derivative is where that swinging settles: a limit, not an approximation.

Derivative Notation #

Three ways to write one idea.

One gradient function, three ways to write it — f'(x), $\frac{dy}{dx}$dy/dx and $\frac{d}{dx}$d/dx

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The gradient of $y=x^2$y = x² is itself a function, 2x. That function needs a name.

Lagrange’s mark: f'(x), read f prime — the derivative of the function f.

Leibniz’s fraction: $\frac{dy}{dx}$dy/dx — a tiny rise over a tiny run, the limit the chords reached.

And $\frac{d}{dx}$d/dx on its own is an instruction: differentiate whatever follows it.

Differentiate twice and each style keeps count: f''(x), or $\frac{d^{2}y}{d{x}^2}$d²y/dx².

Differentiating From First Principles #

Expand, divide by h, then let h vanish.

Expanding the difference quotient and letting h vanish gives the derivative

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Every first-principles derivation starts here: the rise over a step h, divided by h.

Write the difference out in full before dividing by anything.

Only now let h vanish. Dividing by h first, then letting h go to 0, is the method.

Where a Derivative Fails to Exist #

A corner, a cusp, a vertical tangent, a break.

A derivative needs a smooth unbroken point, so differentiable forces continuous

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The two sides of this corner have gradients −1 and +1, so no single value fits.

A cusp is sharper still: the gradients on the two sides run to $-\infty$−∞ and $+\infty$+∞.

$y=\sqrt[3]{x}$y = ∛x stands vertical at 0, and a vertical tangent has no gradient value.

Across a jump, the chords grow steeper without limit as they shrink, so no derivative exists.

So differentiating needs an unbroken curve: differentiable forces continuous.

The reverse fails — a corner is unbroken and still has no gradient there.

Differentiating Linear Functions #

A straight line has one gradient everywhere.

A straight line has one gradient everywhere, so its derivative is a constant

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y = 3x + 1 climbs 3 for every 1 across, at every single point along it.

Its derivative is 3: the constant lifts the line without tilting it.

A flat line has gradient 0, which is why the derivative of a constant is 0.

The Power Rule #

Multiply by the power and drop the power by one.

The same two moves carry the power rule to negative powers and to roots

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First principles gave 2x and $3x^2$3x². Both times the power came down, then dropped.

The power comes down to the front and multiplies. The exponent has not changed yet.

Now the power itself drops by one. Two moves, in that order: 3x squared.

Test a negative power: first principles takes $x^{-1}$x⁻¹ to $-x^{-2}$−x⁻², by the same two moves.

So negative powers take the same two moves: the power comes down to the front, then drops by one.

Roots are powers in disguise: $\sqrt{x}$√x is x to the half, and the same two moves run.

The Constant Multiple and Sum Rules #

Term by term, with constants along for the ride.

Differentiate a sum term by term, and a constant coefficient is left untouched

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The constant multiple rule: differentiate $x^3$x³, then multiply by the coefficient.

The sum rule: differentiate each term separately; the terms do not affect each other.

A constant term differentiates to zero: its graph is horizontal.

The Product Rule #

Differentiate each factor in turn, then add.

Differentiate each factor in turn and add the two results

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Grow a u by v rectangle a little and it gains a strip along each side.

Multiply out: two strips and a corner. The corner vanishes, leaving two terms.

The derivative of a product is never just the product of the two derivatives.

The Quotient Rule #

The same idea, over the denominator squared.

The product rule with a subtraction, all over the denominator squared

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It is the product rule in disguise: write u over v as u times 1 over v.

The product rule needs the derivative of $1/v$1 / v. Write V for v(x + h) and take the difference quotient as $h\to 0$h → 0.

Over the common denominator $v^2$v² that reads u'v − uv', all over the denominator squared.

Unlike the product rule, swapping the two terms changes the sign of the answer.

The Chain Rule #

A function inside a function multiplies rates.

A function inside a function multiplies the two rates of change together

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Read the bracket as a single expression: everything inside travels together.

Now the power rule runs on that object: the power comes down to the front, then drops by one.

The inside changes 3 times as fast as x, so a gradient in u is 3 times steeper in x.

Finally multiply by the inside’s own derivative, which is 3. That gives 15.

Differentiating Trigonometric Functions #

Sine to cosine, and round again in four.

Sine differentiates to cosine, and the pattern repeats after four steps

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Where sine is steepest cosine peaks, and where sine peaks cosine crosses zero.

Four differentiations bring you back to where you started: one closed cycle.

The cycle needs radians: in degrees every derivative picks up a factor of 0.0175.

Differentiating tan x and the Reciprocal Ratios #

The quotient rule turns sin over cos into ${\sec }^2$sec².

The quotient rule turns tan into sec squared, and the same moves cover the reciprocal ratios

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tan x is sin x over cos x — a quotient, so the quotient rule runs on it.

The quotient rule leaves ${\cos }^{2}x+{\sin }^{2}x$cos²x + sin²x as the numerator, over ${\cos }^{2}x$cos²x.

The numerator is 1 by Pythagoras, and 1 over ${\cos }^{2}x$cos²x has a name: ${\sec }^{2}x$sec²x.

sec x is cos x to the −1, so the chain rule gives sin over ${\cos }^2$cos² — sec x tan x.

The same moves run for cosec and cot, and each co-function carries a minus.

The Number E #

The one base whose curve is its own gradient.

e is the one base whose exponential curve has a gradient equal to its own height

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$y=2^x$y = 2ˣ climbs, but its gradient is a little less than its height at every point.

$y=3^x$y = 3ˣ overshoots: its gradient is greater than its height. So try a base between 2 and 3.

Somewhere near 2.718 the two match exactly. That number is called e.

Differentiating Exponentials and Logarithms #

E to the x differentiates to itself.

e to the x differentiates to itself, and the natural log differentiates to one over x

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By the definition of e, $e^x$eˣ is its own derivative — only $e^x$eˣ and its multiples are.

Undo the log: e to the y is x, so differentiating both sides gives 1 over x.

The graphs agree: ln x flattens as x grows, and 1 over x falls toward 0 with it.

With the chain rule, the k inside the exponent comes out to the front as a factor.

Differentiating aˣ and logₐ x #

Every base is e in disguise.

Any exponential differentiates to itself times the log of its base

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Any positive base is e in disguise: $a^x$aˣ is e to the x ln a.

Differentiate as a chain: the inside x ln a changes at rate ln a.

Write it back with base a: the derivative of $a^x$aˣ is $a^x$aˣ times ln a.

ln 2 is less than 1, so the gradient of $2^x$2ˣ is less than its height — $2^x$2ˣ climbs more slowly than $e^x$eˣ.

Check with a = e: ln e is 1, and $e^x$eˣ being its own derivative comes back.

Change of base turns ${\log }_{a}x$logₐ x into a multiple of ln x: its gradient is 1 over x ln a.

Implicit Differentiation #

Differentiate both sides and attach $\frac{dy}{dx}$dy/dx.

Implicit differentiation differentiates both sides, attaching $\frac{dy}{dx}$dy/dx to every y

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$x^2+y^2=25$x² + y² = 25 is a circle, and no rearranging writes y as a single function of x.

Differentiate y-terms as usual, then multiply by $\frac{dy}{dx}$dy/dx — that is the chain rule.

Now collect the $\frac{dy}{dx}$dy/dx terms and solve for it like any other unknown.

The Derivative of an Inverse Function #

Reflect in y = x and the gradient turns over.

Reflecting a graph in y = x turns each gradient into its reciprocal

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Start with $y=\frac{x^2}{2}$y = x²/2, and put the mirror line y = x beside it.

Reflect in that line and the inverse appears: the same curve with x and y swapped.

The reflection swaps the rise with the run, so every gradient becomes its reciprocal.

So when f(a) = b, ${\left(f^{-1}\right)}^{\prime }\left(b\right)$(f⁻¹)'(b) is $1/f^{\prime }\left(a\right)$1 / f'(a) — read f' at a, never at b.

It follows from the chain rule: undoing then doing returns x, whose derivative is 1.

With f(3) = 7 and f'(3) = 5, the inverse has gradient $\frac{1}{5}$1/5 at the point 7.

Differentiating sin⁻¹ x and tan⁻¹ x #

Undo the inverse, then differentiate both sides.

Undoing the inverse and differentiating both sides gives the inverse trig derivatives

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Undo the inverse first: $y={\sin }^{-1}x$y = sin⁻¹ x says exactly that sin y = x.

Differentiate both sides: cos y times $\frac{dy}{dx}$dy/dx equals 1. Divide both sides by cos y.

The answer should be in terms of x: Pythagoras replaces cos y with $\sqrt{1-x^2}$√(1 − x²).

The same three moves on tan y = x give 1 over $1+x^2$1 + x², using $1+{\tan }^{2}y={\sec }^{2}y$1 + tan²y = sec²y.

The graphs agree: ${\tan }^{-1}x$tan⁻¹ x flattens as x grows, and 1 over $1+x^2$1 + x² falls toward 0.

Higher Derivatives #

How fast the gradient itself is changing.

Differentiating a derivative measures how fast the gradient itself is changing

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The gradient of this curve is 2x, which is itself changing as you move along.

Differentiating again gives 2: the gradient grows at a steady rate.

A positive second derivative means the curve bends upwards, like a bowl.