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  2. AP Calculus
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Formula for average rate of change of f(x)f(x)f(x) over [a,b][a, b][a,b]?

f(b)−f(a)b−a\frac{f(b) - f(a)}{b - a}b−af(b)−f(a)​

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Formula for average rate of change of f(x)f(x)f(x) over [a,b][a, b][a,b]?

f(b)−f(a)b−a\frac{f(b) - f(a)}{b - a}b−af(b)−f(a)​

Formula for instantaneous rate of change of f(x)f(x)f(x) at x=cx = cx=c?

f′(c)=lim⁡h→0f(c+h)−f(c)hf'(c) = \lim_{h \to 0} \frac{f(c + h) - f(c)}{h}f′(c)=limh→0​hf(c+h)−f(c)​

What is the power rule for derivatives?

If f(x)=xnf(x) = x^nf(x)=xn, then f′(x)=nxn−1f'(x) = nx^{n-1}f′(x)=nxn−1

What is the constant multiple rule for derivatives?

If f(x)=cf(x)f(x) = cf(x)f(x)=cf(x), then f′(x)=cf′(x)f'(x) = cf'(x)f′(x)=cf′(x)

What is the sum rule for derivatives?

If h(x)=f(x)+g(x)h(x) = f(x) + g(x)h(x)=f(x)+g(x), then h′(x)=f′(x)+g′(x)h'(x) = f'(x) + g'(x)h′(x)=f′(x)+g′(x)

Formula for the derivative of x2x^2x2?

If f(x)=x2f(x) = x^2f(x)=x2, then f′(x)=2xf'(x) = 2xf′(x)=2x

Formula for the derivative of x\sqrt{x}x​?

If f(x)=xf(x) = \sqrt{x}f(x)=x​, then f′(x)=12xf'(x) = \frac{1}{2\sqrt{x}}f′(x)=2x​1​

Formula for the derivative of a constant ccc?

If f(x)=cf(x) = cf(x)=c, then f′(x)=0f'(x) = 0f′(x)=0

What is the point-slope form of a line?

y−y1=m(x−x1)y - y_1 = m(x - x_1)y−y1​=m(x−x1​)

What is the slope-intercept form of a line?

y=mx+by = mx + by=mx+b

How to find the average rate of change of f(x)=x3f(x) = x^3f(x)=x3 on [0,2][0, 2][0,2]?

  1. Calculate f(2)f(2)f(2) and f(0)f(0)f(0). 2. Apply the formula: f(2)−f(0)2−0\frac{f(2) - f(0)}{2 - 0}2−0f(2)−f(0)​. 3. Simplify to get the answer.

How to find the instantaneous rate of change of f(x)=3x2f(x) = 3x^2f(x)=3x2 at x=1x = 1x=1 using the limit definition?

  1. Set up the limit: lim⁡h→0f(1+h)−f(1)h\lim_{h \to 0} \frac{f(1 + h) - f(1)}{h}limh→0​hf(1+h)−f(1)​. 2. Substitute f(x)=3x2f(x) = 3x^2f(x)=3x2. 3. Simplify and evaluate the limit.

Steps to find the equation of the tangent line to f(x)f(x)f(x) at x=ax=ax=a?

  1. Find f(a)f(a)f(a). 2. Find f′(x)f'(x)f′(x). 3. Find f′(a)f'(a)f′(a). 4. Use point-slope form: y−f(a)=f′(a)(x−a)y - f(a) = f'(a)(x - a)y−f(a)=f′(a)(x−a).

How to determine where a function has a horizontal tangent line?

  1. Find the derivative f′(x)f'(x)f′(x). 2. Set f′(x)=0f'(x) = 0f′(x)=0. 3. Solve for xxx.

How to find the average velocity of a particle given its position function s(t)s(t)s(t) over an interval [a,b][a, b][a,b]?

  1. Calculate s(b)s(b)s(b) and s(a)s(a)s(a). 2. Apply the formula: s(b)−s(a)b−a\frac{s(b) - s(a)}{b - a}b−as(b)−s(a)​. 3. Simplify to get the answer.

How to determine if a function is increasing or decreasing at a point?

  1. Find the derivative f′(x)f'(x)f′(x). 2. Evaluate f′(x)f'(x)f′(x) at the point. 3. If f′(x)>0f'(x) > 0f′(x)>0, increasing; if f′(x)<0f'(x) < 0f′(x)<0, decreasing.

How to approximate the instantaneous rate of change using average rate of change?

  1. Choose a small interval around the point. 2. Calculate the average rate of change over that interval. 3. This is an approximation of the instantaneous rate of change.

How to solve for the limit definition of a derivative?

  1. Substitute the function into the limit definition. 2. Simplify the numerator. 3. Cancel out the h term in the denominator. 4. Evaluate the limit.

How to find the instantaneous velocity at t=1 if s(t)=t2+3ts(t) = t^2 + 3ts(t)=t2+3t?

  1. Find the derivative s′(t)=2t+3s'(t) = 2t + 3s′(t)=2t+3. 2. Substitute t=1 into s′(t)s'(t)s′(t). 3. s′(1)=2(1)+3=5s'(1) = 2(1) + 3 = 5s′(1)=2(1)+3=5.

How to find the average rate of change of f(x)=x2+1f(x) = x^2 + 1f(x)=x2+1 from x = 0 to x = 2?

  1. Find f(2)=22+1=5f(2) = 2^2 + 1 = 5f(2)=22+1=5. 2. Find f(0)=02+1=1f(0) = 0^2 + 1 = 1f(0)=02+1=1. 3. Apply formula: 5−12−0=2\frac{5-1}{2-0} = 22−05−1​=2.

Difference between average and instantaneous rate of change?

Average: Over an interval, slope of secant line | Instantaneous: At a point, slope of tangent line.

Compare secant and tangent lines.

Secant: Intersects curve at two points | Tangent: Touches curve at one point, represents derivative.

Compare average and instantaneous velocity.

Average: Change in position over a time interval | Instantaneous: Velocity at a specific time.

Compare the use of the slope formula and the derivative.

Slope formula: for secant lines (average rate of change) | Derivative: for tangent lines (instantaneous rate of change).

Compare the calculation of average vs. instantaneous rate of change.

Average: f(b)−f(a)b−a\frac{f(b)-f(a)}{b-a}b−af(b)−f(a)​ | Instantaneous: lim⁡h→0f(c+h)−f(c)h\lim_{h \to 0} \frac{f(c+h)-f(c)}{h}limh→0​hf(c+h)−f(c)​

What is the difference between average and instantaneous acceleration?

Average acceleration: Change in velocity over a time interval | Instantaneous acceleration: Acceleration at a specific time.

Compare the uses of secant and tangent lines for approximation.

Secant lines: Approximate function behavior over an interval | Tangent lines: Approximate function behavior near a point.

Compare the limit definition and derivative rules.

Limit definition: Fundamental definition, used for proof | Derivative rules: Shortcuts for finding derivatives.

Compare the average rate of change and the Mean Value Theorem.

Average rate of change: Slope of secant line | Mean Value Theorem: Guarantees a point where instantaneous rate of change equals average rate of change.

Compare the applications of average and instantaneous rates of change in physics.

Average rate of change: Used for overall motion analysis | Instantaneous rate of change: Used for specific moment analysis.