Step-by-Step Guidance for Higher Derivatives and Applications (Calculus)

Q1. For $f(x) = x^4 - 3x^3 + 2x - 5$, find $f''(x)$ and $f'''(x)$.

Background

Topic: Higher Derivatives of Polynomial Functions

This question tests your ability to compute second and third derivatives of a polynomial function using basic differentiation rules.

Key Terms and Formulas:

• Derivative: The rate of change of a function.

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

• Second Derivative: $f''(x) = \frac{d^2 f}{dx^2}$

• Third Derivative: $f'''(x) = \frac{d^3 f}{dx^3}$

Step-by-Step Guidance

1. Start by finding the first derivative $f'(x)$ using the power rule for each term.

2. Once you have $f'(x)$, differentiate again to find $f''(x)$.

3. Apply the power rule to each term in $f''(x)$ to find $f'''(x)$.

4. Write out the expressions for $f''(x)$ and $f'''(x)$, but stop before simplifying the final numeric coefficients.

Try solving on your own before revealing the answer!

Final Answer:

$f'(x) = 4x^3 - 9x^2 + 2$

$f''(x) = 12x^2 - 18x$

$f'''(x) = 24x - 18$

Each derivative is found by applying the power rule to each term. The second and third derivatives show how the rate of change itself changes as $x$ increases.

Q2. For $h(x) = 2x^5 - x^4 + 3x^2 - 7$, find $h'''(x)$.

Background

Topic: Higher Derivatives of Polynomial Functions

This question asks you to compute the third derivative of a polynomial function, reinforcing your understanding of repeated differentiation.

Key Terms and Formulas:

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

• Third Derivative: $h'''(x) = \frac{d^3 h}{dx^3}$

Step-by-Step Guidance

1. Find $h'(x)$ by differentiating each term in $h(x)$.

2. Differentiate $h'(x)$ to get $h''(x)$.

3. Differentiate $h''(x)$ to obtain $h'''(x)$.

4. Write out the terms for $h'''(x)$, but leave the coefficients for the student to compute.

Try solving on your own before revealing the answer!

Final Answer:

$h'(x) = 10x^4 - 4x^3 + 6x$

$h''(x) = 40x^3 - 12x^2 + 6$

$h'''(x) = 120x^2 - 24x$

The third derivative shows how the rate of change of the rate of change evolves for this function.

Q3. For $g(x) = \frac{x^4 - 2x^2}{x}$, find $g''(x)$.

Background

Topic: Higher Derivatives of Rational Functions

This question tests your ability to simplify a rational function and then compute its second derivative.

Key Terms and Formulas:

• Quotient Rule: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$

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

• Second Derivative: $g''(x) = \frac{d^2 g}{dx^2}$

Step-by-Step Guidance

1. Simplify $g(x)$ by dividing each term in the numerator by $x$.

2. Rewrite $g(x)$ as a sum of powers of $x$.

3. Find $g'(x)$ using the power rule.

4. Differentiate $g'(x)$ to find $g''(x)$, but stop before simplifying the final expression.

Try solving on your own before revealing the answer!

Final Answer:

$g(x) = x^3 - 2x$

$g'(x) = 3x^2 - 2$

$g''(x) = 6x$

After simplifying, the function becomes a polynomial, making the derivatives straightforward to compute.

Q4. For $y = (x^2 + 1)(x^2 - 1)$, find $\frac{d^2 y}{dx^2}$.

Background

Topic: Second Derivative of a Product of Functions

This question tests your ability to use the product rule and then compute the second derivative.

Key Terms and Formulas:

• Product Rule: $\frac{d}{dx}[u v] = u' v + u v'$

• Second Derivative: $\frac{d^2 y}{dx^2}$

Step-by-Step Guidance

1. Expand $y$ to simplify the expression, or apply the product rule directly.

2. Find $\frac{d y}{dx}$ using the product rule or by differentiating the expanded form.

3. Differentiate again to find $\frac{d^2 y}{dx^2}$, but stop before combining like terms.

Try solving on your own before revealing the answer!

Final Answer:

$y = x^4 - 1$

$\frac{d y}{dx} = 4x^3$

$\frac{d^2 y}{dx^2} = 12x^2$

Expanding the product simplifies the differentiation process.

Q5. For $f(x) = x^6 - 2x^3 + x$, find $f^{(4)}(x)$.

Background

Topic: Fourth Derivative of a Polynomial Function

This question tests your ability to compute higher-order derivatives by repeated application of the power rule.

Key Terms and Formulas:

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

• Fourth Derivative: $f^{(4)}(x) = \frac{d^4 f}{dx^4}$

Step-by-Step Guidance

1. Find $f'(x)$ by differentiating each term.

2. Continue differentiating to find $f''(x)$, $f'''(x)$, and then $f^{(4)}(x)$.

3. Write out the terms for $f^{(4)}(x)$, but leave the coefficients for the student to compute.

Try solving on your own before revealing the answer!

Final Answer:

$f'(x) = 6x^5 - 6x^2 + 1$

$f''(x) = 30x^4 - 12x$

$f'''(x) = 120x^3 - 12$

$f^{(4)}(x) = 360x^2$

The fourth derivative is found by repeated application of the power rule.

Q6. A company’s cost function is $C(x) = 0.01x^3 - 0.3x^2 + 5x + 100$. (a) Find the marginal cost function $C'(x)$. (b) Find $C''(x)$ and interpret its meaning in the context of production costs. (c) At what production level is $C''(x) = 0$? What does this tell us about the cost function?

Background

Topic: Applications of Higher Derivatives in Business

This question tests your ability to find first and second derivatives of a cost function and interpret their meaning in a business context.

Key Terms and Formulas:

• Marginal Cost: $C'(x)$, the rate of change of cost with respect to production.

• Second Derivative: $C''(x)$, the rate of change of marginal cost.

Step-by-Step Guidance

1. Find $C'(x)$ by differentiating each term in $C(x)$.

2. Find $C''(x)$ by differentiating $C'(x)$.

3. Set $C''(x) = 0$ and solve for $x$ to find the production level where the rate of change of marginal cost is zero.

4. Interpret what $C''(x) = 0$ means in terms of the cost function (e.g., inflection point).

Try solving on your own before revealing the answer!

Final Answer:

$C'(x) = 0.03x^2 - 0.6x + 5$

$C''(x) = 0.06x - 0.6$

Setting $C''(x) = 0$ gives $x = 10$.

At $x = 10$, the rate of change of marginal cost is zero, indicating a possible inflection point in the cost function.

Q7. The revenue function for a product is $R(x) = 50x - 0.02x^3$. (a) Find $R''(x)$. (b) Determine where $R''(x) > 0$ and where $R''(x) < 0$. (c) What does the sign of $R''(x)$ tell us about the revenue function’s concavity?

Background

Topic: Concavity and Second Derivative in Business Applications

This question tests your ability to compute the second derivative and interpret its sign in terms of concavity for a revenue function.

Key Terms and Formulas:

• Second Derivative: $R''(x)$

• Concavity: If $R''(x) > 0$, the function is concave up; if $R''(x) < 0$, concave down.

Step-by-Step Guidance

1. Find $R'(x)$ by differentiating $R(x)$.

2. Find $R''(x)$ by differentiating $R'(x)$.

3. Set $R''(x) > 0$ and $R''(x) < 0$ to determine intervals for $x$.

4. Explain what the sign of $R''(x)$ means for the shape of the revenue function.

Try solving on your own before revealing the answer!

Final Answer:

$R'(x) = 50 - 0.06x^2$

$R''(x) = -0.12x$

$R''(x) > 0$ when $x < 0$, $R''(x) < 0$ when $x > 0$.

The sign of $R''(x)$ tells us about the concavity: positive means concave up, negative means concave down.

Q8. A population model for a city is $P(t) = 50000 + 2000t^2 - 10t^3$, where $t$ is years after 2020. (a) Find $P'(t)$ (the rate of population change). (b) Find $P''(t)$ (the acceleration of population change). (c) In what year will the population growth rate start to decrease? (Hint: When does $P''(t) = 0$?)

Background

Topic: Higher Derivatives in Population Models

This question tests your ability to compute first and second derivatives and interpret their meaning in a real-world context.

Key Terms and Formulas:

• First Derivative: $P'(t)$, rate of change of population.

• Second Derivative: $P''(t)$, acceleration of population change.

Step-by-Step Guidance

1. Find $P'(t)$ by differentiating each term in $P(t)$.

2. Find $P''(t)$ by differentiating $P'(t)$.

3. Set $P''(t) = 0$ and solve for $t$ to find when the growth rate starts to decrease.

4. Convert $t$ to the actual year by adding to 2020.

Try solving on your own before revealing the answer!

Final Answer:

$P'(t) = 4000t - 30t^2$

$P''(t) = 4000 - 60t$

Setting $P''(t) = 0$ gives $t = 66.67$, so the year is $2020 + 66.67 = 2086.67$ (about 2087).

The population growth rate starts to decrease in the year 2087.

Q9. Find all derivatives of $f(x) = x^4 - 8x^3 + 12x^2$ until you reach zero.

Background

Topic: Successive Derivatives of a Polynomial

This question tests your ability to repeatedly differentiate a polynomial until the derivative becomes zero.

Key Terms and Formulas:

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

Step-by-Step Guidance

1. Find $f'(x)$ by differentiating each term.

2. Continue differentiating to find $f''(x)$, $f'''(x)$, etc.

3. Stop when the derivative becomes zero.

Try solving on your own before revealing the answer!

Final Answer:

$f'(x) = 4x^3 - 24x^2 + 24x$

$f''(x) = 12x^2 - 48x + 24$

$f'''(x) = 24x - 48$

$f^{(4)}(x) = 24$

$f^{(5)}(x) = 0$

After the fourth derivative, the function becomes constant, and the fifth derivative is zero.

Q10. If $f(x) = ax^3 + bx^2 + cx + d$ and $f''(2) = 12$, $f'''(1) = 18$, find the values of $a$ and $b$.

Background

Topic: Using Derivative Values to Solve for Coefficients

This question tests your ability to use information about higher derivatives at specific points to solve for unknown coefficients in a polynomial.

Key Terms and Formulas:

• Second Derivative: $f''(x) = 6a x + 2b$

• Third Derivative: $f'''(x) = 6a$

Step-by-Step Guidance

1. Write expressions for $f''(x)$ and $f'''(x)$ in terms of $a$ and $b$.

2. Plug $x = 2$ into $f''(x)$ and set equal to 12.

3. Plug $x = 1$ into $f'''(x)$ and set equal to 18.

4. Set up the system of equations for $a$ and $b$, but stop before solving.

Try solving on your own before revealing the answer!

Final Answer:

$f''(x) = 6a x + 2b$

$f''(2) = 12 \Rightarrow 12a + 2b = 12$

$f'''(x) = 6a$

$f'''(1) = 18 \Rightarrow 6a = 18 \Rightarrow a = 3$

Substitute $a = 3$ into the first equation: $12(3) + 2b = 12 \Rightarrow 36 + 2b = 12 \Rightarrow 2b = -24 \Rightarrow b = -12$

So $a = 3$, $b = -12$.

Q11. A manufacturing company finds that their efficiency function is $E(t) = -t^4 + 8t^3 - 18t^2 + 12t + 50$, where $t$ is hours worked in a day. (a) Find when the rate of efficiency change is maximized (find where $E''(t) = 0$). (b) Determine the concavity of the efficiency function at $t = 2$ hours.

Background

Topic: Higher Derivatives and Concavity in Business Contexts

This question tests your ability to compute second derivatives and interpret their meaning for optimization and concavity.

Key Terms and Formulas:

• Second Derivative: $E''(t)$

• Concavity: If $E''(t) > 0$, function is concave up; if $E''(t) < 0$, concave down.

Step-by-Step Guidance

1. Find $E'(t)$ by differentiating $E(t)$.

2. Find $E''(t)$ by differentiating $E'(t)$.

3. Set $E''(t) = 0$ and solve for $t$ to find when the rate of efficiency change is maximized.

4. Evaluate $E''(2)$ to determine concavity at $t = 2$ hours.

Try solving on your own before revealing the answer!

Final Answer:

$E'(t) = -4t^3 + 24t^2 - 36t + 12$

$E''(t) = -12t^2 + 48t - 36$

Setting $E''(t) = 0$ gives $-12t^2 + 48t - 36 = 0$; solving yields $t = 1$ and $t = 3$.

At $t = 2$, $E''(2) = -12(4) + 48(2) - 36 = -48 + 96 - 36 = 12$; since $E''(2) > 0$, the function is concave up at $t = 2$ hours.