Tasks: Differentiation Rules & Tangent Lines

Session 05-03 Practice Problems

1 Problem 1: Basic Differentiation Rules (x)

Find the derivatives of the following functions using the appropriate rules:

  1. \(f(x) = 5x^4 - 3x^2 + 7x - 2\)

  2. \(g(x) = 2\sqrt{x} + \frac{3}{x} - \frac{1}{x^3}\)

  3. \(h(x) = \frac{1}{2}x^{-4} + 4x^{1/3}\)

a) Apply power rule and sum rule: \[f'(x) = 20x^3 - 6x + 7\]

b) Rewrite using exponents: \(g(x) = 2x^{1/2} + 3x^{-1} - x^{-3}\) \[g'(x) = 2 \cdot \frac{1}{2}x^{-1/2} + 3(-x^{-2}) - (-3x^{-4})\] \[= x^{-1/2} - 3x^{-2} + 3x^{-4} = \frac{1}{\sqrt{x}} - \frac{3}{x^2} + \frac{3}{x^4}\]

c) Apply power rule: \[h'(x) = \frac{1}{2}(-4x^{-5}) + 4 \cdot \frac{1}{3}x^{-2/3}\] \[= -2x^{-5} + \frac{4}{3}x^{-2/3} = -\frac{2}{x^5} + \frac{4}{3\sqrt[3]{x^2}}\]

2 Problem 2: Product Rule Practice (x)

Differentiate the following functions using the product rule:

  1. \(f(x) = (3x^2 + 1)(x - 4)\)

  2. \(g(x) = x^3(2x^2 - 5x + 1)\)

  3. Verify your answer to part (a) by first expanding, then differentiating.

a) Using product rule with \(u = 3x^2 + 1\) and \(v = x - 4\):

  • \(u' = 6x\)
  • \(v' = 1\)

\[f'(x) = (6x)(x - 4) + (3x^2 + 1)(1)\] \[= 6x^2 - 24x + 3x^2 + 1 = 9x^2 - 24x + 1\]

b) Using product rule with \(u = x^3\) and \(v = 2x^2 - 5x + 1\):

  • \(u' = 3x^2\)
  • \(v' = 4x - 5\)

\[g'(x) = (3x^2)(2x^2 - 5x + 1) + (x^3)(4x - 5)\] \[= 6x^4 - 15x^3 + 3x^2 + 4x^4 - 5x^3\] \[= 10x^4 - 20x^3 + 3x^2\]

c) Expand first: \(f(x) = (3x^2 + 1)(x - 4) = 3x^3 - 12x^2 + x - 4\)

Then differentiate: \(f'(x) = 9x^2 - 24x + 1\) ✓ (matches part a)

3 Problem 3: Quotient Rule Applications (xx)

Find the derivatives using the quotient rule:

  1. \(f(x) = \frac{x^2 + 3}{x - 1}\)

  2. \(g(x) = \frac{2x - 5}{3x + 2}\)

  3. \(h(x) = \frac{x^3 - 1}{x^2 + 1}\)

a) Using quotient rule with \(u = x^2 + 3\) and \(v = x - 1\):

  • \(u' = 2x\)
  • \(v' = 1\)

\[f'(x) = \frac{(2x)(x - 1) - (x^2 + 3)(1)}{(x - 1)^2}\] \[= \frac{2x^2 - 2x - x^2 - 3}{(x - 1)^2} = \frac{x^2 - 2x - 3}{(x - 1)^2}\]

Factoring numerator: \(\frac{(x - 3)(x + 1)}{(x - 1)^2}\)

b) Using quotient rule with \(u = 2x - 5\) and \(v = 3x + 2\):

  • \(u' = 2\)
  • \(v' = 3\)

\[g'(x) = \frac{(2)(3x + 2) - (2x - 5)(3)}{(3x + 2)^2}\] \[= \frac{6x + 4 - 6x + 15}{(3x + 2)^2} = \frac{19}{(3x + 2)^2}\]

c) Using quotient rule with \(u = x^3 - 1\) and \(v = x^2 + 1\):

  • \(u' = 3x^2\)
  • \(v' = 2x\)

\[h'(x) = \frac{(3x^2)(x^2 + 1) - (x^3 - 1)(2x)}{(x^2 + 1)^2}\] \[= \frac{3x^4 + 3x^2 - 2x^4 + 2x}{(x^2 + 1)^2} = \frac{x^4 + 3x^2 + 2x}{(x^2 + 1)^2}\]

Factor numerator: \(\frac{x(x^3 + 3x + 2)}{(x^2 + 1)^2}\)

4 Problem 4: Finding Tangent Lines (xx)

For each function, find the equation of the tangent line at the given point:

  1. \(f(x) = x^3 - 2x^2 + 1\) at \(x = 2\)

  2. \(g(x) = \frac{x + 1}{x - 1}\) at \(x = 3\)

  3. \(h(x) = x^2(x - 3)\) at the point where \(x = 1\)

a) For \(f(x) = x^3 - 2x^2 + 1\) at \(x = 2\):

Step 1: Find the point \[f(2) = 8 - 8 + 1 = 1\] Point: \((2, 1)\)

Step 2: Find the slope \[f'(x) = 3x^2 - 4x\] \[f'(2) = 12 - 8 = 4\]

Step 3: Equation \[y - 1 = 4(x - 2)\] \[y = 4x - 7\]

b) For \(g(x) = \frac{x + 1}{x - 1}\) at \(x = 3\):

Step 1: Find the point \[g(3) = \frac{4}{2} = 2\] Point: \((3, 2)\)

Step 2: Find the slope using quotient rule \[g'(x) = \frac{(1)(x - 1) - (x + 1)(1)}{(x - 1)^2} = \frac{x - 1 - x - 1}{(x - 1)^2} = \frac{-2}{(x - 1)^2}\] \[g'(3) = \frac{-2}{4} = -\frac{1}{2}\]

Step 3: Equation \[y - 2 = -\frac{1}{2}(x - 3)\] \[y = -\frac{1}{2}x + \frac{7}{2}\]

c) For \(h(x) = x^2(x - 3)\) at \(x = 1\):

Step 1: Expand or use product rule Expanding: \(h(x) = x^3 - 3x^2\), so \(h(1) = 1 - 3 = -2\) Point: \((1, -2)\)

Step 2: Find the slope \[h'(x) = 3x^2 - 6x\] \[h'(1) = 3 - 6 = -3\]

Step 3: Equation \[y - (-2) = -3(x - 1)\] \[y = -3x + 1\]

Three graphs are shown side by side. (a) The cubic f(x) = x^3 - 2x^2 + 1 with tangent line y = 4x - 7 touching at point (2, 1). (b) The rational function g(x) = (x+1)/(x-1) with tangent line y = -0.5x + 3.5 touching at point (3, 2), showing the decreasing hyperbolic curve. (c) The cubic h(x) = x^2(x-3) with tangent line y = -3x + 1 touching at point (1, -2).

5 Problem 5: Linear Approximation in Business (xx)

A company’s profit function is: \[P(x) = -0.2x^2 + 12x - 50\] where \(x\) is the number of items sold (in thousands) and \(P\) is profit in thousands of dollars.

  1. Find the profit when \(x = 25\) thousand items.

  2. Find the marginal profit function \(P'(x)\) and evaluate it at \(x = 25\).

  3. Use linear approximation to estimate the profit when \(x = 26\) thousand items.

  4. Calculate the actual profit at \(x = 26\) and compare with your estimate. What is the error?

  5. Interpret the marginal profit at \(x = 25\) in business terms.

a) Current profit: \[P(25) = -0.2(625) + 12(25) - 50 = -125 + 300 - 50 = 125\] The profit is \(\$125{,}000\).

b) Marginal profit: \[P'(x) = -0.4x + 12\] \[P'(25) = -0.4(25) + 12 = -10 + 12 = 2\]

c) Linear approximation: \[P(26) \approx P(25) + P'(25) \cdot (26 - 25)\] \[\approx 125 + 2(1) = 127\] Estimated profit: \(\$127{,}000\)

d) Actual profit: \[P(26) = -0.2(676) + 12(26) - 50 = -135.2 + 312 - 50 = 126.8\] Actual profit: \(\$126{,}800\)

Error: \(|127 - 126.8| = 0.2\) thousand = \(\$200\)

The approximation is very accurate (error is only 0.16%).

e) Business interpretation:

The marginal profit \(P'(25) = 2\) means that when producing 25,000 items, selling one additional thousand items will increase profit by approximately \(\$2{,}000\). Since this is positive, the company should consider increasing production beyond 25,000 units.

The graph shows the downward-opening parabola P(x) = -0.2x^2 + 12x - 50 with a tangent line (linear approximation) drawn at x = 25. The current point (25, 125) is marked, along with the actual value at x = 26 (126.8) and the estimated value (127). A dotted vertical line between the two points at x = 26 shows the small approximation error of 0.2.

6 Problem 6: Optimization with Marginal Analysis (xxx)

A manufacturer has the following functions:

  • Cost: \(C(x) = 10{,}000 + 50x + 0.5x^2\)
  • Revenue: \(R(x) = 200x - 0.5x^2\)

where \(x\) is the number of units produced and sold.

  1. Find the profit function \(P(x) = R(x) - C(x)\).

  2. Find the marginal cost, marginal revenue, and marginal profit functions.

  3. Determine the production level where marginal revenue equals marginal cost.

  4. Verify that this is the same production level where marginal profit equals zero.

  5. Calculate the actual profit at this optimal production level.

  6. Create a graph showing all three marginal functions on the same axes.

a) Profit function:

\[P(x) = R(x) - C(x) = (200x - 0.5x^2) - (10{,}000 + 50x + 0.5x^2)\] \[P(x) = -x^2 + 150x - 10{,}000\]

b) Marginal functions:

Marginal cost: \[MC(x) = C'(x) = 50 + x\]

Marginal revenue: \[MR(x) = R'(x) = 200 - x\]

Marginal profit: \[MP(x) = P'(x) = -2x + 150\]

c) Set MR = MC:

\[200 - x = 50 + x\] \[150 = 2x\] \[x = 75 \text{ units}\]

d) Check where MP = 0:

\[-2x + 150 = 0\] \[x = 75 \text{ units}\] ✓

This confirms the fundamental principle: profit is maximized where MR = MC.

e) Optimal profit:

\[P(75) = -(75)^2 + 150(75) - 10{,}000\] \[= -5{,}625 + 11{,}250 - 10{,}000 = -4{,}375\]

Since \(P(x) = -x^2 + 150x - 10{,}000\) is a downward-opening parabola (coefficient of \(x^2\) is negative), \(x = 75\) is indeed the maximum.

However, the maximum profit is still negative. This means:

  • The company loses money at all production levels
  • At \(x = 75\), losses are minimized (not maximized profit)
  • The company should consider if fixed costs are too high

f) Graph of marginal functions:

The graph shows three linear functions: MC(x) = 50 + x (increasing), MR(x) = 200 - x (decreasing), and MP(x) = -2x + 150 (decreasing). MC and MR intersect at x = 75 (marked with a point and dashed vertical line), which is labeled as the optimal production level. At the same x = 75, the marginal profit line crosses zero, confirming the MR = MC principle.

Key observations:

  1. MR = MC at x = 75: This is where profit is optimized (losses minimized)
  2. MP = 0 at x = 75: Confirms this is a critical point
  3. For x < 75: MR > MC, so increasing production increases profit
  4. For x > 75: MR < MC, so increasing production decreases profit

7 Problem 7: Sensitivity and Error Analysis (xxx)

A pharmaceutical company uses the formula \(D(t) = \frac{100t}{t^2 + 4}\) to model the concentration of a drug in the bloodstream (in mg/L) \(t\) hours after administration.

  1. Find \(D'(t)\) using the quotient rule.

  2. Evaluate \(D(2)\) and \(D'(2)\). Interpret both values.

  3. Use linear approximation to estimate \(D(2.1)\).

  4. The angle of inclination of the tangent line at \(t = 2\) is \(\alpha = \arctan(D'(2))\). Calculate \(\alpha\) in degrees.

  5. At what time \(t\) is the rate of change of drug concentration equal to zero? What is the concentration at that time?

  6. Create a graph showing the concentration function and the tangent line at \(t = 2\).

a) Derivative using quotient rule:

Let \(u = 100t\) and \(v = t^2 + 4\)

  • \(u' = 100\)
  • \(v' = 2t\)

\[D'(t) = \frac{100(t^2 + 4) - 100t(2t)}{(t^2 + 4)^2}\] \[= \frac{100t^2 + 400 - 200t^2}{(t^2 + 4)^2}\] \[= \frac{-100t^2 + 400}{(t^2 + 4)^2}\] \[= \frac{100(4 - t^2)}{(t^2 + 4)^2}\]

b) Evaluate at t = 2:

Concentration: \[D(2) = \frac{100(2)}{4 + 4} = \frac{200}{8} = 25 \text{ mg/L}\]

Rate of change: \[D'(2) = \frac{100(4 - 4)}{64} = 0 \text{ mg/L per hour}\]

Interpretation:

  • At \(t = 2\) hours, the drug concentration is 25 mg/L
  • At this moment, the concentration is neither increasing nor decreasing (instantaneous rate = 0)
  • This suggests \(t = 2\) is a critical point (likely the maximum concentration)

c) Linear approximation:

\[D(2.1) \approx D(2) + D'(2)(2.1 - 2) = 25 + 0(0.1) = 25 \text{ mg/L}\]

Since \(D'(2) = 0\), the tangent line is horizontal, so the approximation gives the same value.

Let’s verify with the actual value: \[D(2.1) = \frac{100(2.1)}{(2.1)^2 + 4} = \frac{210}{4.41 + 4} = \frac{210}{8.41} \approx 24.97 \text{ mg/L}\]

The approximation is very accurate!

d) Angle of inclination:

\[\alpha = \arctan(D'(2)) = \arctan(0) = 0°\]

The tangent line is horizontal at \(t = 2\).

e) When is D’(t) = 0?

\[\frac{100(4 - t^2)}{(t^2 + 4)^2} = 0\]

The numerator must equal zero: \[100(4 - t^2) = 0\] \[4 - t^2 = 0\] \[t^2 = 4\] \[t = \pm 2\]

Since \(t\) represents time after administration, we take \(t = 2\) hours.

Concentration at this time: \[D(2) = 25 \text{ mg/L}\]

This is the maximum concentration (the peak drug level).

f) Graph:

The graph shows the drug concentration D(t) = 100t/(t^2 + 4) as a curve that rises from 0, reaches a peak of 25 mg/L at t = 2 hours, then gradually decreases. A horizontal dashed tangent line at y = 25 touches the curve at the peak, confirming D’(2) = 0. A dotted vertical line at t = 2 marks the peak level.

Medical interpretation:

  • The drug reaches peak concentration (25 mg/L) at \(t = 2\) hours
  • Before 2 hours: concentration is increasing (\(D'(t) > 0\))
  • After 2 hours: concentration is decreasing (\(D'(t) < 0\))
  • The horizontal tangent at \(t = 2\) confirms this is the maximum
  • This information is crucial for determining optimal dosing schedules