MathClario
← All lessons
High school∫Calculus15 min read
πŸ“‰

Derivatives: Complete Lesson with Animated Tangent and Key Formulas

Instantaneous rate, tangent line, derivative function. Every exam-level formula, with an interactive tangent visualization.

Prerequisites :

  • Functions and graphs
  • Algebra
  • Limits (basic idea)

The idea in one sentence

A derivative measures a function’s local slope β€” how ff changes β€œjust next to” a point.

Equivalently: it’s the slope of the tangent line to the curve at that point.

See it: the tangent in action

Slide the cursor to move the point along the curve. The red tangent follows, and its slope updates live:

Point
β€”
Pente de la tangente (nombre dΓ©rivΓ©)
β€”

Things to notice:

  • When x0<0x_0 < 0 β†’ slope is negative (curve going down)
  • When x0=0x_0 = 0 β†’ slope is zero (bottom of parabola)
  • When x0>0x_0 > 0 β†’ slope is positive (curve going up)

Formula check: fβ€²(x)=2xf'(x) = 2x. At x0=1x_0 = 1, slope = 2. At x0=βˆ’2x_0 = -2, slope = βˆ’4-4.

Rigorous definition

The derivative of ff at aa, written fβ€²(a)f'(a), is the limit (if it exists):

fβ€²(a)=lim⁑hβ†’0f(a+h)βˆ’f(a)hf'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}

The quotient f(a+h)βˆ’f(a)h\dfrac{f(a + h) - f(a)}{h} is the difference quotient between aa and a+ha+h: the slope of the secant line through those two points. As hβ†’0h \to 0, the secant tends to the tangent.

Equation of the tangent line

The tangent to ff at x=ax = a has equation:

y=fβ€²(a)β‹…(xβˆ’a)+f(a)y = f'(a) \cdot (x - a) + f(a)

Example: For f(x)=x2f(x) = x^2 at a=3a = 3:

  • f(3)=9f(3) = 9
  • fβ€²(3)=6f'(3) = 6
  • Tangent: y=6(xβˆ’3)+9=6xβˆ’9y = 6(x - 3) + 9 = 6x - 9.

Standard derivative formulas

Memorize these. Everything builds on them.

f(x)f(x)fβ€²(x)f'(x)Domain
kk (constant)00R\mathbb{R}
xx11R\mathbb{R}
xnx^n (n∈Zn \in \mathbb{Z})nβ‹…xnβˆ’1n \cdot x^{n-1}R\mathbb{R} (or Rβˆ—\mathbb{R}^* if n<0n < 0)
x\sqrt{x}12x\dfrac{1}{2\sqrt{x}}x>0x > 0
1x\dfrac{1}{x}βˆ’1x2-\dfrac{1}{x^2}xβ‰ 0x \neq 0
ln⁑(x)\ln(x)1x\dfrac{1}{x}x>0x > 0
exe^xexe^xR\mathbb{R}
sin⁑(x)\sin(x)cos⁑(x)\cos(x)R\mathbb{R}
cos⁑(x)\cos(x)βˆ’sin⁑(x)-\sin(x)R\mathbb{R}

Derivative operations

OperationFormula
Sum(u+v)β€²=uβ€²+vβ€²(u + v)' = u' + v'
Constant Γ— function(kβ‹…u)β€²=kβ‹…uβ€²(k \cdot u)' = k \cdot u'
Product(uv)β€²=uβ€²v+uvβ€²(uv)' = u'v + uv'
Quotient(uv)β€²=uβ€²vβˆ’uvβ€²v2\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}
Chain rule(v∘u)β€²=uβ€²β‹…(vβ€²βˆ˜u)(v \circ u)' = u' \cdot (v' \circ u)

Examples

Example 1 (sum): f(x)=3x2βˆ’5x+7f(x) = 3x^2 - 5x + 7

fβ€²(x)=6xβˆ’5f'(x) = 6x - 5.

Example 2 (product): f(x)=x2β‹…exf(x) = x^2 \cdot e^x

fβ€²(x)=2xβ‹…ex+x2β‹…ex=ex(2x+x2)f'(x) = 2x \cdot e^x + x^2 \cdot e^x = e^x(2x + x^2).

Example 3 (quotient): f(x)=xx2+1f(x) = \dfrac{x}{x^2 + 1}

fβ€²(x)=1β‹…(x2+1)βˆ’xβ‹…2x(x2+1)2=1βˆ’x2(x2+1)2f'(x) = \dfrac{1 \cdot (x^2 + 1) - x \cdot 2x}{(x^2 + 1)^2} = \dfrac{1 - x^2}{(x^2 + 1)^2}.

Example 4 (chain): f(x)=sin⁑(3x+1)f(x) = \sin(3x + 1)

fβ€²(x)=3β‹…cos⁑(3x+1)f'(x) = 3 \cdot \cos(3x + 1).

Key applications

1. Monotonicity

fβ€²>0f' > 0 on an interval β‡’ ff is increasing. fβ€²<0f' < 0 β‡’ decreasing. fβ€²=0f' = 0 at an isolated point β‡’ local extremum (confirm with sign table).

2. Finding an extremum

f(x)=βˆ’x2+4x+1f(x) = -x^2 + 4x + 1.

  • fβ€²(x)=βˆ’2x+4f'(x) = -2x + 4.
  • fβ€²=0β€…β€ŠβŸΊβ€…β€Šx=2f' = 0 \iff x = 2.
  • Sign change (+ then βˆ’) β†’ max at 2, value f(2)=5f(2) = 5.
-112345-11234567maxf(x) = -xΒ² + 4x + 1

3. Tangent equation

See section above.

Common mistakes

  1. Forgetting uβ€²u' in a chain rule: (sin⁑(2x))β€²=2cos⁑(2x)(\sin(2x))' = 2\cos(2x), NOT cos⁑(2x)\cos(2x).
  2. Confusing product and composition: (uv)β€²β‰ (v(u))β€²(uv)' \neq (v(u))'.
  3. Sign in quotient rule: it’s uβ€²vβˆ’uvβ€²u'v - uv' (minus).
  4. Derivative of a constant is 0, not the constant.

Solved exercises

Exercise 1

Compute the derivative of f(x)=(x2+1)3f(x) = (x^2 + 1)^3.

Show solution

Chain rule: u=x2+1u = x^2 + 1, uβ€²=2xu' = 2x. Outer: v(x)=x3v(x) = x^3, vβ€²(x)=3x2v'(x) = 3x^2. fβ€²(x)=2xβ‹…3(x2+1)2=6x(x2+1)2f'(x) = 2x \cdot 3(x^2 + 1)^2 = 6x(x^2 + 1)^2.

Exercise 2

Find the tangent to f(x)=exf(x) = e^x at x=0x = 0.

Show solution

f(0)=1f(0) = 1, fβ€²(0)=1f'(0) = 1. Tangent: y=x+1y = x + 1.

Exercise 3

Study the monotonicity of f(x)=x3βˆ’3x+2f(x) = x^3 - 3x + 2 on R\mathbb{R}.

Show solution

fβ€²(x)=3(xβˆ’1)(x+1)f'(x) = 3(x-1)(x+1). Positive for x<βˆ’1x < -1, negative on (βˆ’1,1)(-1, 1), positive for x>1x > 1. So ff is increasing on (βˆ’βˆž,βˆ’1](-\infty, -1], decreasing on [βˆ’1,1][-1, 1], increasing on [1,+∞)[1, +\infty). Local max at βˆ’1-1: f(βˆ’1)=4f(-1) = 4. Local min at 11: f(1)=0f(1) = 0.

Takeaways

  • fβ€²(a)f'(a) = slope of the tangent at aa.
  • Tangent: y=fβ€²(a)(xβˆ’a)+f(a)y = f'(a)(x - a) + f(a).
  • Monotonicity = sign of the derivative.
  • Local extremum β‡’ fβ€²=0f' = 0 (converse not always).
  • Master the 9 standard formulas and 5 operations.

Going further

#derivative#tangent#instantaneous rate#grade 12#AP#lesson