Math - Calculus

Derivatives Explained Simply

A derivative answers one question: how fast is something changing right now? Not on average over an hour, but at a single instant. On a graph that means: how steep is the curve at exactly this point?

f′(x) = lim_h→0 [ f(x+h) − f(x) ] / h The derivative at x is the slope of the tangent line — the limit of the secant slopes as the second point moves closer

From secant to tangent

Finding the slope of a straight line is easy: rise over run. A curve has a different slope at every point, so we start with an approximation. Take the point P you care about and a second point Q a little to the right. The straight line through both is called a secant, and its slope is easy to compute — but it only describes the average behaviour between P and Q.

Now slide Q towards P. The gap, usually called h, gets smaller and smaller, and the secant tips over into a particular position: the tangent at P. Its slope is the instantaneous rate of change we were looking for. We cannot simply set h to zero, because that would put a zero in the denominator — hence the detour through a limit.

As Q slides towards P, the secant tips into the tangent. Its slope is the derivative at P.

Power rule: f(x) = xⁿ → f′(x) = n · xⁿ⁻¹

f(x) = x³ → f′(x) = 3x²
f(x) = 5x² → f′(x) = 10x
f(x) = 7 → f′(x) = 0

f(x) = x³ − 3x + 4 → f′(x) = 3x² − 3
Slope at x = 2: f′(2) = 12 − 3 = 9

The differentiation rules
Rule	Formula	Example
Power rule	xⁿ → n·xⁿ⁻¹	x⁴ → 4x³
Constant factor	c·f → c·f′	6x² → 12x
Sum rule	f + g → f′ + g′	x² + x → 2x + 1
Product rule	f·g → f′g + fg′	x·sin x → sin x + x·cos x
Chain rule	f(g(x)) → f′(g(x))·g′(x)	(2x+1)³ → 3(2x+1)²·2

What it is used for

Where f′(x) = 0 the tangent is horizontal, which marks a possible maximum or minimum. The second derivative tells you which: negative means the curve bends downwards, so it is a maximum; positive means a minimum. That is the core of every optimisation problem — the largest area for a given fence, the least material for a given volume.

The physical reading matters just as much. Differentiate distance with respect to time and you get speed. Differentiate speed again and you get acceleration. A car's speedometer displays nothing but the first derivative of its distance function.

Common mistakes

Forgetting the inner derivative in the chain rule. (2x+1)³ becomes 3(2x+1)² times 2.
Treating every f′(x) = 0 as an extremum. For f(x) = x³ the derivative vanishes at 0, but that point is a saddle.
Carrying the constant along. x² + 5 differentiates to 2x, not 2x + 5.

Practice problem

For f(x) = x³ − 6x² + 9x, find every extremum and state whether it is a maximum or a minimum.

Show solution

f′(x) = 3x² − 12x + 9 = 0 → x² − 4x + 3 = 0 → x = 1 and x = 3.
f″(x) = 6x − 12. f″(1) = −6 → maximum at x = 1, f(1) = 4. f″(3) = 6 → minimum at x = 3, f(3) = 0.