Derivatives

What a derivative is

The derivative measures how fast a function changes at a single instant. It is the slope of a secant line taken to the limit as the two points merge, which turns an average rate into an instantaneous one:

\[\class{var-fn}{f}'(x) = \lim_{h \to 0}\frac{\class{var-fn}{f}(x+h) - \class{var-fn}{f}(x)}{h}\]

Notation

\(f'(x)\) Lagrange notation, compact and best for stating rules

\(\dfrac{dy}{dx}\) Leibniz notation, names the variables and survives the chain rule

\(\dfrac{d}{dx}\big[\ \big]\) the operator form, read as "differentiate this with respect to \(x\)"

\(f''(x)\), \(\dfrac{d^2y}{dx^2}\) the second derivative, the rate at which the rate changes

\(f'(a)\) a single number: the slope at that one point

What it tells you

Geometrically the slope of the tangent line at \(x\)

Physically an instantaneous rate: position gives velocity, velocity gives acceleration

Tangent line \(y - f(a) = f'(a)(x - a)\), straight from point-slope form

Differentiable \(\Rightarrow\) continuous but not the reverse one-way implication

No derivative at corners, cusps, vertical tangents, and any break in the graph not differentiable

Running that limit by hand is slow, so it gets done once per function family. Everything below is the result of that work, packaged into rules you can apply directly.

The power rule and its companions

Any term that is a variable raised to a constant power follows one pattern: drop the exponent in front, then reduce it by one. It holds for every real exponent, negative and fractional included:

\[\frac{d}{dx}\,x^{\class{var-gn}{n}} = \class{var-gn}{n}\,x^{\class{var-gn}{n}-1}\]

Rules that come with it

Constant \(\dfrac{d}{dx}c = 0\); a flat line has no slope

Constant multiple \(\dfrac{d}{dx}\,c\,f(x) = c\,f'(x)\); coefficients ride along

Sum and difference differentiate term by term, in any order

Linear term \(\dfrac{d}{dx}x = 1\), the power rule with \(n = 1\)

Watch out \(\dfrac{d}{dx}b^x \neq x\,b^{x-1}\); a constant base is a different rule entirely common error

Rewrite before differentiating

\(\sqrt{x}\) \(= x^{1/2}\), so the derivative is \(\tfrac{1}{2}x^{-1/2}\)

\(\dfrac{1}{x^3}\) \(= x^{-3}\), giving \(-3x^{-4}\)

\(\dfrac{x^2 + 5x}{x}\) split it into \(x + 5\) first and skip the quotient rule saves work

\(x^2 \cdot x^4\) combine to \(x^6\) rather than reaching for the product rule

Habit simplify into powers first; it removes most of the hard rules essential habit

Most messy-looking derivatives are just algebra that was never cleaned up. Rewriting first is faster than differentiating and simplifying afterward.

Product and quotient rules

Derivatives do not distribute across multiplication. When two functions are multiplied, each one changes while the other is still present, so both contributions are counted:

\[(\class{var-fn}{u}\class{var-gn}{v})' = \class{var-fn}{u}'\class{var-gn}{v} + \class{var-fn}{u}\class{var-gn}{v}'\]

Division follows from the same idea, with the second term subtracted and the denominator squared:

\[\left(\frac{\class{var-fn}{u}}{\class{var-gn}{v}}\right)' = \frac{\class{var-fn}{u}'\class{var-gn}{v} - \class{var-fn}{u}\class{var-gn}{v}'}{\class{var-gn}{v}^2}\]

Using them

Product order addition, so the order of the two terms does not matter

Quotient order subtraction, so the numerator's derivative must come first sign error trap

Constant denominator use the constant multiple rule instead; no quotient rule needed

Three factors \((uvw)' = u'vw + uv'w + uvw'\), one term per factor

Never \((uv)' \neq u'v'\) common error

Example: \(y = x^3\ln x\)

Label the pieces \(u = x^3\), \(v = \ln x\)

Their derivatives \(u' = 3x^2\), \(v' = \dfrac{1}{x}\)

Assemble \(y' = 3x^2\ln x + x^3 \cdot \dfrac{1}{x}\)

Simplify \(y' = 3x^2\ln x + x^2 = x^2(3\ln x + 1)\) answer

A quotient can always be rewritten as a product with a negative exponent, so the quotient rule is a convenience rather than a necessity. Use whichever leaves less algebra behind.

The chain rule

When one function is nested inside another, the rates multiply. Differentiate the outer function while leaving the inner one untouched, then multiply by the derivative of the inner:

\[\frac{d}{dx}\,\class{var-fn}{f}\big(\class{var-gn}{g}(x)\big) = \class{var-fn}{f}'\big(\class{var-gn}{g}(x)\big) \cdot \class{var-gn}{g}'(x)\]

Procedure

1. Identify the layers the outer function is whatever you would evaluate last

2. Differentiate outward apply the outer rule with the inner expression left in place

3. Multiply by the inside tack on \(g'(x)\); skipping this is the most common mistake in calculus never skip

More layers keep multiplying inward: \(f'(g(h))\cdot g'(h)\cdot h'\)

Leibniz form \(\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}\), where the \(du\) appears to cancel

Patterns worth recognizing

\([g(x)]^n\) \(n[g(x)]^{n-1}g'(x)\)

\(e^{g(x)}\) \(e^{g(x)}g'(x)\)

\(\ln g(x)\) \(\dfrac{g'(x)}{g(x)}\)

\(\sin g(x)\) \(\cos\big(g(x)\big)\,g'(x)\)

\(\sin^2 x\) read as \((\sin x)^2\), giving \(2\sin x\cos x\) notation trap

Think of gears in mesh. If the outer wheel turns three times as fast as the middle one, and the middle turns twice as fast as the inner, the outer turns six times as fast as the inner. The chain rule is that multiplication.

Derivatives of common functions

These are the building blocks every problem is assembled from. Each one assumes a bare \(x\) inside; with anything else in there, the chain rule adds a factor of the inside's derivative.

Trigonometric

\(\sin x\) \(\cos x\)

\(\cos x\) \(-\sin x\)

\(\tan x\) \(\sec^2 x\)

\(\sec x\) \(\sec x \tan x\)

\(\csc x\) \(-\csc x \cot x\)

\(\cot x\) \(-\csc^2 x\)

Pattern every co-function derivative carries a minus sign memory hook

Exponential and logarithmic

\(e^x\) \(e^x\), the function that is its own derivative unique

\(b^x\) \(b^x \ln b\)

\(\ln x\) \(\dfrac{1}{x}\), for \(x > 0\)

\(\log_b x\) \(\dfrac{1}{x \ln b}\)

\(\ln|x|\) \(\dfrac{1}{x}\) on both sides of zero matters for integrals

Inverse trigonometric

\(\arcsin x\) \(\dfrac{1}{\sqrt{1 - x^2}}\)

\(\arccos x\) \(-\dfrac{1}{\sqrt{1 - x^2}}\)

\(\arctan x\) \(\dfrac{1}{1 + x^2}\)

\(\operatorname{arccot} x\) \(-\dfrac{1}{1 + x^2}\)

Worth noticing no trig functions appear in the answers, which is why they show up in integrals

Applying the chain rule to these

\(\cos(5x)\) \(-5\sin(5x)\)

\(e^{-2x}\) \(-2e^{-2x}\)

\(\ln(x^2 + 1)\) \(\dfrac{2x}{x^2 + 1}\)

\(\tan(3x^2)\) \(6x\sec^2(3x^2)\)

Radians only the trig derivatives are false in degrees common error

Learn these as a table, not as derivations. Fluency here is what makes the harder rules feel mechanical rather than difficult.

Implicit differentiation

Some curves cannot be solved for \(y\), and others are painful to solve. Instead, differentiate both sides of the equation as they stand, treating \(y\) as an unknown function of \(x\) so every \(y\) term picks up a chain rule factor:

\[\frac{d}{dx}\,\class{var-gn}{y}^{\,n} = n\,\class{var-gn}{y}^{\,n-1}\,\frac{d\class{var-gn}{y}}{dx}\]

Procedure

1. Differentiate both sides with respect to \(x\), term by term

2. Tag every \(y\) each \(y\) term produces a \(\dfrac{dy}{dx}\) factor never skip

3. Mixed terms a term like \(xy\) needs the product rule as well

4. Collect move every \(\dfrac{dy}{dx}\) term to one side, everything else to the other

5. Factor and divide the answer usually contains both \(x\) and \(y\) expected

Example: \(x^2 + xy + y^2 = 7\)

Differentiate \(2x + \left(y + x\dfrac{dy}{dx}\right) + 2y\dfrac{dy}{dx} = 0\)

Collect \((x + 2y)\dfrac{dy}{dx} = -(2x + y)\)

Solve \(\dfrac{dy}{dx} = -\dfrac{2x + y}{x + 2y}\) answer

Use it substitute a point on the curve to get that tangent's slope

Horizontal tangents where the numerator is zero and the denominator is not

The same idea drives related rates: differentiate a relationship with respect to time instead of \(x\), and every variable contributes its own rate. It also derives the inverse trig derivatives above, starting from \(\sin y = x\).

Key takeaways

Summary

Definition \(f'(x) = \lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}\), the slope of the tangent line

Power rule \(\frac{d}{dx}x^n = nx^{n-1}\); rewrite roots and reciprocals as powers first

Product rule \((uv)' = u'v + uv'\), one term per factor

Quotient rule \(\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\), numerator's derivative first

Chain rule \(\frac{d}{dx}f(g(x)) = f'(g(x))g'(x)\); the inner derivative is the step people forget

Core table \(\sin \to \cos\), \(\cos \to -\sin\), \(e^x \to e^x\), \(\ln x \to \frac{1}{x}\)

Implicit differentiate through, attach \(\frac{dy}{dx}\) to every \(y\), then solve for it

Choosing a rule simplify first; nested means chain, multiplied means product