Calculus AB: Derivatives - Definitions, Rules, and Applications

Chapter 2.1 - Limit Definition of the Derivative vs. Alternate Definition of the Derivative (Day 3 of Unit)

I. Definition of the Derivative

A. General Definition ($f'(x)$)

• The derivative of a function $f(x)$, denoted as $f'(x)$, is defined using a limit:
  $f'(x) = \lim_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}$. (Historically, $h$ is often used instead of $\Delta x$).

• This definition is used to find a new function, $f'(x)$, which represents the slope of the tangent line drawn to the graph of $f(x)$ at any given point $x$.

B. Alternate Form of the Definition (Derivative at a specific point $x=a$)

• The instantaneous rate of change or the slope of the tangent line to the graph of $f(x)$ at a specific point $x=a$ is given by:
  $f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$.

• This form evaluates to a numerical value, representing the slope of the tangent line at that particular point $a$. It is a specific instance of the general derivative at a single point.

C. Examples of Using the Limit Definition

1. Finding $f'(x)$ for $f(x) = \frac{x}{x+2}$ using the general definition:
   $f'(x) = \lim<em>{\Delta x \to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}$ $= \lim</em>{\Delta x \to 0} \frac{\frac{x+\Delta x}{(x+\Delta x)+2} - \frac{x}{x+2}}{\Delta x}$

   • Find a common denominator for the numerator:
     $= \lim_{\Delta x \to 0} \frac{(x+\Delta x)(x+2) - x(x+\Delta x+2)}{\Delta x (x+\Delta x+2)(x+2)}$

   • Expand the numerator:
     $= \lim_{\Delta x \to 0} \frac{x^2+2x+x\Delta x+2\Delta x - x^2-x\Delta x-2x}{\Delta x (x+\Delta x+2)(x+2)}$

   • Simplify the numerator (terms cancel out):
     $= \lim_{\Delta x \to 0} \frac{2\Delta x}{\Delta x (x+\Delta x+2)(x+2)}$

   • Cancel out $\Delta x$ (since $\Delta x \ne 0$ as $\Delta x \to 0$):
     $= \lim_{\Delta x \to 0} \frac{2}{(x+\Delta x+2)(x+2)}$

   • Substitute $\Delta x = 0$:
     $= \frac{2}{(x+0+2)(x+2)} = \frac{2}{(x+2)^2}$

   • Therefore, $f'(x) = \frac{2}{(x+2)^2}$.

2. Finding $f'(-1)$ for $f(x) = \frac{2}{x+2}$ using the alternate definition ($a = -1$):

   • First, find $f(a) = f(-1)$:
     $f(-1) = \frac{2}{-1+2} = \frac{2}{1} = 2$

   • Apply the alternate definition:
     $f'(-1) = \lim<em>{x \to -1} \frac{f(x) - f(-1)}{x - (-1)} = \lim</em>{x \to -1} \frac{\frac{2}{x+2} - 2}{x+1}$

   • Find a common denominator for the numerator:
     $= \lim<em>{x \to -1} \frac{\frac{2 - 2(x+2)}{x+2}}{x+1} = \lim</em>{x \to -1} \frac{2 - 2x - 4}{(x+2)(x+1)}$

   • Simplify the numerator:
     $= \lim<em>{x \to -1} \frac{-2x - 2}{(x+2)(x+1)} = \lim</em>{x \to -1} \frac{-2(x+1)}{(x+2)(x+1)}$

   • Cancel out $(x+1)$ (since $x \ne -1$ as $x \to -1$):
     $= \lim_{x \to -1} \frac{-2}{x+2}$

   • Substitute $x = -1$:
     $= \frac{-2}{-1+2} = \frac{-2}{1} = -2$

   • Therefore, $f'(-1) = -2$. (This matches what we would get by evaluating $f'(x) = \frac{-2}{(x+2)^2}$ at $x=-1$: $\frac{-2}{(-1+2)^2} = \frac{-2}{1^2} = -2$).

Section 2.2 - Differentiation Rules and Rates of Change

II. Basic Differentiation Rules

• These rules provide shortcuts for finding derivatives, avoiding the lengthy limit definition process.

1. Constant Rule:

   • The derivative of a constant function is zero.

   • $\frac{d}{dx}[c] = 0$

   • Example: $f(x) = 3 \implies f'(x) = 0$. (A horizontal line has a slope of zero).

2. Power Rule:

   • To differentiate a power function $x^n$ (where $n$ is any real number), bring the exponent down as a coefficient and subtract $1$ from the exponent.

   • $\frac{d}{dx}[x^n] = n \cdot x^{n-1}$

   • Example: $f(x) = x^5 \implies f'(x) = 5x^{5-1} = 5x^4$.

3. Constant Multiple Rule:

   • When a function is multiplied by a constant, the derivative is the constant times the derivative of the function.

   • $\frac{d}{dx}[c \cdot f(x)] = c \cdot f'(x)$

   • Example: $f(x) = 3x^4 \implies f'(x) = 3 \cdot (4x^3) = 12x^3$.

4. Sum/Difference Rule:

   • The derivative of a sum or difference of two functions is the sum or difference of their derivatives.

   • $\frac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x)$

5. Derivatives of Sine and Cosine (Memorize!):

   • $\frac{d}{dx}[\sin x] = \cos x$

   • $\frac{d}{dx}[\cos x] = -\sin x$

6. Derivative of the Natural Exponential Function:

   • The derivative of $e^x$ is $e^x$ itself.

   • $\frac{d}{dx}[e^x] = e^x$

III. Examples of Applying Differentiation Rules

A. Basic Differentiation

1. $f(x) = -2 \implies f'(x) = 0$ (Constant Rule).

2. $g(x) = 3x \implies g'(x) = 3$ (Constant Multiple and Power Rule ($x^1$)).

3. $y = 2x^3 - x^2 + 3x \implies y' = 6x^2 - 2x + 3$ (Sum/Difference, Constant Multiple, Power Rules).

B. Mixed Functions

1. $y = 5 + \sin x \implies \frac{dy}{dx} = 0 + \cos x = \cos x$ (Constant and Sine Rules).

2. $g(t) = \pi \cos t \implies g'(t) = \pi \cdot (-\sin t) = -\pi \sin t$ (Constant Multiple and Cosine Rules).

3. $f(x) = e^3 \implies f'(x) = 0$ (Since $e^3$ is a constant).

4. $h(x) = e^x - 4x \implies h'(x) = e^x - 4$ (Exponential, Power, Difference Rules).

5. $g(x) = 7e^x - \cos x \implies g'(x) = 7e^x - (-\sin x) = 7e^x + \sin x$ (Constant Multiple, Exponential, Cosine, Difference Rules).

6. $g(x) = 7e^5 \implies g'(x) = 0$ (Since $7e^5$ is a constant).

C. Functions Requiring Rewriting

1. $y = \frac{2}{3x^2}$:

   • Rewrite as $y = \frac{2}{3} x^{-2}$

   • $y' = \frac{2}{3}(-2)x^{-2-1} = -\frac{4}{3}x^{-3} = -\frac{4}{3x^3}$.

2. $y = \frac{\pi}{(3x)^2}$:

   • Rewrite as $y = \frac{\pi}{9x^2} = \frac{\pi}{9} x^{-2}$

   • $y' = \frac{\pi}{9}(-2)x^{-2-1} = -\frac{2\pi}{9}x^{-3} = -\frac{2\pi}{9x^3}$.

3. $h(x) = \frac{2x^3+x}{x^2}$:

   • Rewrite by dividing each term:
     $h(x) = \frac{2x^3}{x^2} + \frac{x}{x^2} = 2x + x^{-1}$

   • $h'(x) = 2(1)x^{1-1} + (-1)x^{-1-1} = 2x^0 - x^{-2} = 2 - \frac{1}{x^2}$.

4. $f(x) = \frac{1}{3\sqrt{x}} + \sqrt[5]{x^2}$ (Interpreted from transcript's calculation):

   • Rewrite using rational exponents:
     $f(x) = \frac{1}{3}x^{-1/2} + x^{2/5}$

   • $f'(x) = \frac{1}{3}(-\frac{1}{2})x^{-1/2-1} + \frac{2}{5}x^{2/5-1}$

   • $f'(x) = -\frac{1}{6}x^{-3/2} + \frac{2}{5}x^{-3/5}$.

IV. Derivative at a Point and Tangent Lines

A. Finding the Derivative at a Specific Point (Example 2)

• Problem: Find the derivative of $y = 3x^5 + 10x^2$ at the point $(1, 13)$.

• Steps:

  1. Find the general derivative $y'$.
     $y' = \frac{d}{dx}[3x^5 + 10x^2] = 15x^4 + 20x$

  2. Evaluate $y'$ at the given $x$-coordinate, $x=1$.
     $y'(1) = 15(1)^4 + 20(1) = 15 + 20 = 35$.

• This value, $35$, is the slope of the tangent line to the curve at the point $(1, 13)$.

• Equation of the Tangent Line (Optional but related): Using the point-slope form $y - y<em>1 = m(x - x</em>1)$:
  $y - 13 = 35(x - 1)$
  $y = 35(x - 1) + 13$

B. Finding Points with Horizontal Tangents (Example 3)

• A horizontal tangent line indicates where the slope of the curve is zero (i.e., the derivative $f'(x) = 0$).

1. Problem A: Find the point(s) where $y = x^2 + 1$ has a horizontal tangent.

   • Find the derivative: $\frac{dy}{dx} = 2x$

   • Set the derivative to zero: $2x = 0 \implies x = 0$

   • Find the corresponding $y$-coordinate: $y(0) = (0)^2 + 1 = 1$

   • The function has a horizontal tangent at $(0, 1)$.

2. Problem B: Find the point(s) where $y = x^3 + x$ has a horizontal tangent.

   • Find the derivative: $\frac{dy}{dx} = 3x^2 + 1$

   • Set the derivative to zero: $3x^2 + 1 = 0 \implies 3x^2 = -1 \implies x^2 = -\frac{1}{3}$

   • Since $x^2$ cannot be negative for real numbers, there are no real values of $x$ for which the derivative is zero.

   • Conclusion: The function $y = x^3 + x$ has no horizontal tangent lines.

V. Rates of Change: AROC, IROC, Position, Velocity, Acceleration

A. Types of Rates of Change

• A rate is a comparison between two quantities of different kinds (e.g., km/hr for speed, points/game for scoring rate).

1. Average Rate of Change (AROC):

   • Represents the slope of the secant line (or chord) connecting two points $(a, f(a))$ and $(b, f(b))$ on a curve.

   • Formula: $\text{AROC} = \frac{\text{change in } y}{\text{change in } x} = \frac{\Delta y}{\Delta x} = \frac{f(b) - f(a)}{b - a}$.

   • No limit is involved. Used for an interval.

2. Instantaneous Rate of Change (IROC):

   • Represents the slope of the tangent line at a single point on a curve.

   • It is the derivative of the function evaluated at that point.

   • Achieved by taking the limit of the AROC as the interval shrinks to zero (i.e., using the definition of the derivative).

B. Kinematics (Motion along a Line)

• This describes the relationship between an object's position, velocity, and acceleration over time.

1. **Position Function ($s(t)$):

   • The original function, often denoted as $s(t)$ (or $h(t), y(t)$), describing the object's location at time $t$.

2. **Velocity Function ($v(t)$):

   • The instantaneous rate of change of position.

   • $v(t) = s'(t)$ (the first derivative of the position function).

   • Units: m/s, ft/s, etc.

   • v(t) > 0 means the object is moving in the positive direction (e.g., rising).

   • v(t) < 0 means the object is moving in the negative direction (e.g., falling).

   • $v(t) = 0$ means the object is momentarily at rest or changing direction.

3. **Acceleration Function ($a(t)$):

   • The instantaneous rate of change of velocity.

   • $a(t) = v'(t) = s''(t)$ (the first derivative of velocity, or the second derivative of position).

   • Units: m/s$, ft/s$, etc.

C. Specific Kinematics Concepts

1. Position Function for Objects Under Gravity:

   • General form: $s(t) = \frac{1}{2}gt^2 + v<em>0t + s</em>0$

     • $s_0$: initial height (at $t=0$).

     • $v_0$: initial velocity (at $t=0$).

     • $g$: acceleration due to gravity.

       • If units are feet per second ($ft/s$): $g = -32 \text{ ft/s}^2$. So, $s(t) = -16t^2 + v<em>0t + s</em>0$.

       • If units are meters per second ($m/s$): $g = -9.8 \text{ m/s}^2$. So, $s(t) = -4.9t^2 + v<em>0t + s</em>0$.

2. Speed:

   • Speed is the magnitude of velocity; it is always non-negative.

   • $\text{Speed} = |v(t)|$.

3. Increasing/Decreasing Speed:

   • Speed is increasing if velocity and acceleration have the same sign (v(t) \cdot a(t) > 0).

   • Speed is decreasing if velocity and acceleration have opposite signs (v(t) \cdot a(t) < 0).

4. Displacement:

   • The net change in position from an initial time $a$ to a final time $b$.

   • $\text{Displacement} = s(b) - s(a)$.

   • Displacement can be positive, negative, or zero.

5. Total Distance Traveled:

   • The sum of the absolute values of the distances traveled over all sub-intervals where the direction of motion does not change.

   • This requires finding all times $t$ where $v(t)=0$ (and thus the object might change direction) within the interval $[a,b]$.

   • Then, sum the absolute values of the displacements for each sub-interval: $|s(t<em>1) - s(a)| + |s(t</em>2) - s(t<em>1)| + \dots + |s(b) - s(t</em>k)|$, where $t<em>1, t</em>2, \dots, t_k$ are the times the object changes direction.

VI. Word Problems Involving Rates of Change

A. Problem 1: Ball Dropped from CN Tower

• Scenario: A ball is dropped from an observation deck 450 m above the ground. ($s<em>0 = 450 ext{ m}$, $v</em>0 = 0 ext{ m/s}$).

1. a. Write the equation of the distance of the ball after $t$ seconds.

   • Since units are meters, use $g = -9.8 \text{ m/s}^2$.

   • $s(t) = -4.9t^2 + v<em>0t + s</em>0$

   • $s(t) = -4.9t^2 + 0t + 450 \implies s(t) = -4.9t^2 + 450$

2. b. Find the average velocity of the ball from $t=3$ seconds to $t=5$ seconds (AROC).

   • $s(5) = -4.9(5)^2 + 450 = -4.9(25) + 450 = -122.5 + 450 = 327.5 \text{ m}$

   • $s(3) = -4.9(3)^2 + 450 = -4.9(9) + 450 = -44.1 + 450 = 405.9 \text{ m}$

   • $\text{AROC} = \frac{s(5) - s(3)}{5 - 3} = \frac{327.5 - 405.9}{2} = \frac{-78.4}{2} = -39.2 \text{ m/s}$

3. c. Find the velocity of the ball after $t=5$ seconds (IROC).

   • First, find the velocity function: $v(t) = s'(t) = \frac{d}{dt}[-4.9t^2 + 450] = -9.8t$

   • Evaluate at $t = 5$:
     $v(5) = -9.8(5) = -49 \text{ m/s}$

4. d. At what time will the ball hit the ground?

   • The ball hits the ground when its height is $0$, so set $s(t) = 0$.

   • $0 = -4.9t^2 + 450$

   • $4.9t^2 = 450$

   • $t^2 = \frac{450}{4.9} \approx 91.8367$

   • $t = \sqrt{91.8367} \approx 9.583 \text{ seconds}$ (Time must be positive).

5. e. With what velocity will the ball hit the ground?

   • Evaluate the velocity function $v(t)$ at the time found in part (d).

   • $v(9.583) = -9.8(9.583) \approx -93.913 \text{ m/s}$

B. Problem 2: Ball Thrown into Air

• Scenario: A ball is thrown into the air with a velocity of 40 ft/s. Its height ($y$) in feet after $t$ seconds is given by $y = -16t^2 + 40t$.

1. a. Find the average velocity for time periods beginning when $t=2$ lasting:

   • First, find the height at $t=2$: $h(2) = -16(2)^2 + 40(2) = -16(4) + 80 = -64 + 80 = 16 \text{ ft}$.

   • i) $0.5$ s (Interval $[2, 2.5]$):

     • $h(2.5) = -16(2.5)^2 + 40(2.5) = -16(6.25) + 100 = -100 + 100 = 0 \text{ ft}$

     • $\text{AROC} = \frac{h(2.5) - h(2)}{2.5 - 2} = \frac{0 - 16}{0.5} = -32 \text{ ft/s}$

   • ii) $0.1$ s (Interval $[2, 2.1]$):

     • $h(2.1) = -16(2.1)^2 + 40(2.1) = -16(4.41) + 84 = -70.56 + 84 = 13.44 \text{ ft}$

     • $\text{AROC} = \frac{h(2.1) - h(2)}{2.1 - 2} = \frac{13.44 - 16}{0.1} = \frac{-2.56}{0.1} = -25.6 \text{ ft/s}$

   • iii) $0.05$ s (Interval $[2, 2.05]$):

     • $h(2.05) = -16(2.05)^2 + 40(2.05) = -16(4.2025) + 82 = -67.24 + 82 = 14.76 \text{ ft}$

     • $\text{AROC} = \frac{h(2.05) - h(2)}{2.05 - 2} = \frac{14.76 - 16}{0.05} = \frac{-1.24}{0.05} = -24.8 \text{ ft/s}$

   • iv) $0.01$ s (Interval $[2, 2.01]$):

     • $h(2.01) = -16(2.01)^2 + 40(2.01) = -16(4.0401) + 80.4 = -64.6416 + 80.4 = 15.7584 \text{ ft}$

     • $\text{AROC} = \frac{h(2.01) - h(2)}{2.01 - 2} = \frac{15.7584 - 16}{0.01} = \frac{-0.2416}{0.01} = -24.16 \text{ ft/s}$

   • Observation: As the time interval shrinks, the average velocities approach -24 ft/s, which suggests the instantaneous velocity at $t=2$ is -24 ft/s.

2. b. Find the instantaneous velocity when $t=2$.

   • Find the velocity function: $v(t) = h'(t) = \frac{d}{dt}[-16t^2 + 40t] = -32t + 40$

   • Evaluate at $t = 2$:
     $v(2) = -32(2) + 40 = -64 + 40 = -24 \text{ ft/s}$

C. Problem 3: Arrow Shot Upward on the Moon

• Scenario: An arrow is shot upward on the moon with a velocity of 58 m/s. Its height ($y$) in meters after $t$ seconds is $y = 58t - 0.83t^2$.

1. a. Find the average velocity over the given time intervals (approximating IROC at $t=1$):

   • First, find the height at $t=1$: $s(1) = 58(1) - 0.83(1)^2 = 58 - 0.83 = 57.17 \text{ m}$.

   • i) $[1, 2]$: $\frac{s(2) - s(1)}{2 - 1} = \frac{(58(2) - 0.83(2)^2) - 57.17}{1} = \frac{116 - 3.32 - 57.17}{1} = \frac{112.68 - 57.17}{1} = 55.51 \text{ m/s}$

   • ii) $[1, 1.5]$: $\frac{s(1.5) - s(1)}{1.5 - 1} = \frac{(58(1.5) - 0.83(1.5)^2) - 57.17}{0.5} = \frac{87 - 1.8675 - 57.17}{0.5} = \frac{27.9625}{0.5} = 55.925 \text{ m/s}$

   • iii) $[1, 1.1]$: $\frac{s(1.1) - s(1)}{1.1 - 1} = \frac{(58(1.1) - 0.83(1.1)^2) - 57.17}{0.1} = \frac{63.8 - 1.0043 - 57.17}{0.1} = \frac{5.6257}{0.1} = 56.257 \text{ m/s}$

   • iv) $[1, 1.01]$: $\frac{s(1.01) - s(1)}{1.01 - 1} = \frac{(58(1.01) - 0.83(1.01)^2) - 57.17}{0.01} = \frac{58.58 - 0.846683 - 57.17}{0.01} = \frac{0.563317}{0.01} = 56.3317 \text{ m/s}$

   • v) $[1, 1.001]$: $\frac{s(1.001) - s(1)}{1.001 - 1} = \frac{(58(1.001) - 0.83(1.001)^2) - 57.17}{0.001} = \frac{58.058 - 0.83166083 - 57.17}{0.001} = \frac{0.05633917}{0.001} = 56.33917 \text{ m/s}$

   • Approximation: The instantaneous velocity at $t=1$ is approximately $56.34 \text{ m/s}$.

2. b. Find the instantaneous velocity after 1s.

   • Find the velocity function: $v(t) = s'(t) = \frac{d}{dt}[58t - 0.83t^2] = 58 - 1.66t$

   • Evaluate at $t = 1$:
     $v(1) = 58 - 1.66(1) = 56.34 \text{ m/s}$

3. c. Is the arrow rising or falling at this time?

   • Since v(1) = 56.34 \text{ m/s} > 0 , the velocity is positive, meaning the arrow is rising at $t=1$ second.

VII. Graphical Interpretation of Rates of Change and Derivatives

A. Interpreting AROC from Graphs

• When viewing a graph of two quantities, the AROC between two points is represented by the slope of the chord (the straight line segment) connecting those points.

1. Example: Mice Colony Population

   • a. Estimate the average rate of increase in population:

     • i. From week 3 to week 6:

       • Read points from graph: $(3, 110)$ and $(6, 240)$.

       • $\text{AROC} = \frac{f(6) - f(3)}{6 - 3} = \frac{240 - 110}{3} = \frac{130}{3} \approx 43.3 \text{ mice/week}$

     • ii. The seven-week period (from week 0 to week 7):

       • Read points from graph: $(0, 50)$ and $(7, 310)$.

       • $\text{AROC} = \frac{f(7) - f(0)}{7 - 0} = \frac{310 - 50}{7} = \frac{260}{7} \approx 37.14 \text{ mice/week}$

   • b. What is the overall trend in population growth? The curve appears to be flattening, indicating that the rate of population growth is slowing down over this period.

2. Example: Ball Bearing Fall Distance

   • Problem: Find the average speed in the time interval $2 \le t \le 4$.

   • Read points from graph: $(2, 20)$ and $(4, 80)$.

   • $\text{Average Speed} = \frac{f(4) - f(2)}{4 - 2} = \frac{80 - 20}{2} = \frac{60}{2} = 30 \text{ m/s}$

B. Interpreting IROC from Graphs

• The instantaneous rate of change at a point $A$ is given by the slope of the tangent line to the curve at point $A$.

• This concept is visualized by letting a second point $B$ move closer and closer to point $A$. The secant line (chord) $AB$ approaches the tangent line at $A$. Consequently, the AROC approaches the IROC.

C. Estimating Derivatives from a Graph:

• The derivative $f'(x)$ at a point $x$ is the slope of the tangent line at that point.

1. $f'(-3) = 0$ (Horizontal tangent at a local maximum).

2. $f'(-1) = -2$ (Estimated slope of the tangent line at $x=-1$. Example tangent points: $( -1, -1 )$ and $( -0.5, -2 )$. Slope $=\frac{-2 - (-1)}{-0.5 - (-1)} = \frac{-1}{0.5} = -2$).

3. $f'(0) = \text{DNE}$ (Does Not Exist - because the graph has a sharp point/corner at $x=0$).

4. $f'(1) = 2$ (Estimated slope of the tangent line at $x=1$. Example tangent points: $( 0.5, -0.5 )$ and $( 1.5, 1.5 )$. Slope $=\frac{1.5 - (-0.5)}{1.5 - 0.5} = \frac{2}{1} = 2$).

5. $f'(4) = 0$ (Horizontal tangent at a local minimum).

VIII. Approximating Derivatives with Tables

• 
  
  

• When only a table of discrete values is provided (and the function is continuous and differentiable), we can approximate the derivative at a point by calculating the AROC between points closest to the desired $x$-value.

• 
  
  

• Given Table for $f(x)$:
  
  

  $x$	$f(x)$
  1	-2
  3	1
  7	5
  9	3

  1. a. Approximate $f'(2)$.

     • Use the points around $x=2$: $(1, -2)$ and $(3, 1)$.

     • $f'(2) \approx \frac{f(3) - f(1)}{3 - 1} = \frac{1 - (-2)}{2} = \frac{3}{2}$.

  2. b. Estimate $f'(5)$.

     • Use the points around $x=5$: $(3, 1)$ and $(7, 5)$.

     • $f'(5) \approx \frac{f(7) - f(3)}{7 - 3} = \frac{5 - 1}{4} = \frac{4}{4} = 1$.

  IX. Application of the Limit Definition of the Derivative

  • Problem: Find the slope of the graph of $f(x) = 2x^2 - 5x$ at the value $x=1$ using the definition of the derivative.

  • Steps:

    1. Apply the general limit definition:
       $f'(x) = \lim<em>{\Delta x \to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}$ $= \lim</em>{\Delta x \to 0} \frac{[2(x+\Delta x)^2 - 5(x+\Delta x)] - [2x^2 - 5x]}{\Delta x}$

    2. Expand $(x+\Delta x)^2$:
       $= \lim_{\Delta x \to 0} \frac{[2(x^2 + 2x\Delta x + (\Delta x)^2) - 5x - 5\Delta x] - [2x^2 - 5x]}{\Delta x}$

    3. Distribute and simplify the numerator:
       $= \lim<em>{\Delta x \to 0} \frac{2x^2 + 4x\Delta x + 2(\Delta x)^2 - 5x - 5\Delta x - 2x^2 + 5x}{\Delta x}$ $= \lim</em>{\Delta x \to 0} \frac{4x\Delta x + 2(\Delta x)^2 - 5\Delta x}{\Delta x}$

    4. Factor out $\Delta x$ from the numerator:
       $= \lim_{\Delta x \to 0} \frac{\Delta x (4x + 2\Delta x - 5)}{\Delta x}$

    5. Cancel $\Delta x$ (since $\Delta x \ne 0$ as $\Delta x \to 0$):
       $= \lim_{\Delta x \to 0} (4x + 2\Delta x - 5)$

    6. Evaluate the limit by substituting $\Delta x = 0$:
       $f'(x) = 4x + 2(0) - 5 = 4x - 5$

    7. Find the slope at $x=1$:
       $f'(1) = 4(1) - 5 = 4 - 5 = -1$

    • The slope of the graph of $f(x) = 2x^2 - 5x$ at $x=1$ is $-1$.