Q1a. Find \( \frac{dy}{dx} \) for the curve defined by \( x^2 + 2y^2 - 2xy + 3x - 3y = 5 \).

Background

Topic: Implicit Differentiation

This question tests your ability to use implicit differentiation to find the derivative \( \frac{dy}{dx} \) when y is defined implicitly in terms of x.

Key Terms and Formulas:

• Implicit differentiation: Differentiating both sides of an equation with respect to x, treating y as a function of x.

• Chain rule: When differentiating terms involving y, use \( \frac{d}{dx}[y] = \frac{dy}{dx} \).

Step-by-Step Guidance

1. Differentiate each term of the equation with respect to x. Remember to use the chain rule for terms involving y.

2. For example, \( \frac{d}{dx}[x^2] = 2x \), \( \frac{d}{dx}[2y^2] = 4y \frac{dy}{dx} \), \( \frac{d}{dx}[-2xy] = -2y - 2x \frac{dy}{dx} \).

3. Continue differentiating the remaining terms: \( \frac{d}{dx}[3x] = 3 \), \( \frac{d}{dx}[-3y] = -3 \frac{dy}{dx} \), \( \frac{d}{dx}[5] = 0 \).

4. Combine all differentiated terms into a single equation.

5. Group all terms containing \( \frac{dy}{dx} \) on one side and all other terms on the other side.

Try solving on your own before revealing the answer!

Q1b. Determine all points (x, y) on the curve where the tangent line is horizontal.

Background

Topic: Tangent Lines and Derivatives

This question tests your understanding of when a tangent line is horizontal, which occurs when \( \frac{dy}{dx} = 0 \).

Key Terms and Formulas:

• Horizontal tangent: \( \frac{dy}{dx} = 0 \)

• Use the derivative found in part (a): \( \frac{dy}{dx} = \frac{2x - 2y + 3}{4y - 2x - 3} \)

Step-by-Step Guidance

1. Set the numerator of the derivative equal to zero: \( 2x - 2y + 3 = 0 \).

2. Solve for y in terms of x: \( 2x + 3 = 2y \Rightarrow y = \frac{2x + 3}{2} \).

3. Substitute this expression for y back into the original curve equation to find the corresponding x values.

4. Solve the resulting equation for x.

5. Once you have x, use the expression for y to find the corresponding y values.

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Q1c. Find the coordinates of all points (x, y) on the curve where the tangent line is vertical.

Background

Topic: Tangent Lines and Derivatives

This question tests your understanding of when a tangent line is vertical, which occurs when the denominator of \( \frac{dy}{dx} \) is zero.

Key Terms and Formulas:

• Vertical tangent: Denominator of \( \frac{dy}{dx} \) is zero.

• Use the derivative from part (a): \( \frac{dy}{dx} = \frac{2x - 2y + 3}{4y - 2x - 3} \)

Step-by-Step Guidance

1. Set the denominator equal to zero: \( 4y - 2x - 3 = 0 \).

2. Solve for y in terms of x: \( 4y = 2x + 3 \Rightarrow y = \frac{2x + 3}{4} \).

3. Substitute this expression for y back into the original curve equation to find the corresponding x values.

4. Solve the resulting equation for x.

5. Once you have x, use the expression for y to find the corresponding y values.

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Q1d. Find an equation of the line tangent to the curve at the point (-5, -1).

Background

Topic: Tangent Line Equation

This question tests your ability to find the equation of a tangent line using the point-slope form and the derivative at a specific point.

Key Terms and Formulas:

• Tangent line: \( y - y_1 = m(x - x_1) \), where m is the slope at \( (x_1, y_1) \).

• Use \( \frac{dy}{dx} \) from part (a) evaluated at \( (-5, -1) \).

Step-by-Step Guidance

1. Plug \( x = -5 \) and \( y = -1 \) into the derivative formula to find the slope m.

2. Calculate the numerator and denominator separately before dividing.

3. Use the point-slope form with \( (x_1, y_1) = (-5, -1) \) and the slope m.

4. Write the equation in the form \( y - (-1) = m(x + 5) \).

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Q2. Given \( f(x) \) is defined by \( x^2 + xy + y^2 = 7 \). Find the value of the second derivative at the point (3, -2).

Background

Topic: Implicit Differentiation and Second Derivatives

This question tests your ability to find the second derivative \( \frac{d^2y}{dx^2} \) when y is defined implicitly in terms of x.

Key Terms and Formulas:

• Implicit differentiation: Differentiate both sides with respect to x.

• Second derivative: Differentiate \( \frac{dy}{dx} \) with respect to x, using the chain rule.

Step-by-Step Guidance

1. Differentiate the equation \( x^2 + xy + y^2 = 7 \) implicitly with respect to x to find \( \frac{dy}{dx} \).

2. Collect terms and solve for \( \frac{dy}{dx} \) in terms of x and y.

3. Differentiate \( \frac{dy}{dx} \) with respect to x to find \( \frac{d^2y}{dx^2} \), using the chain rule for y.

4. Plug in the values x = 3 and y = -2 into your expressions for \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \).

5. Evaluate the second derivative at the given point.

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Q3a. Find \( \frac{dy}{dx} \) for \( e^{xy} = \sec(x) + y^3 \).

Background

Topic: Implicit Differentiation

This question tests your ability to use implicit differentiation for equations involving both x and y.

Key Terms and Formulas:

• Implicit differentiation

• Chain rule

• Product rule for differentiating \( xy \)

Step-by-Step Guidance

1. Differentiate both sides of the equation with respect to x.

2. For \( e^{xy} \), use the chain rule: \( \frac{d}{dx}[e^{xy}] = e^{xy}(y + x \frac{dy}{dx}) \).

3. For \( \sec(x) \), differentiate to get \( \sec(x) \tan(x) \).

4. For \( y^3 \), differentiate to get \( 3y^2 \frac{dy}{dx} \).

5. Collect all terms with \( \frac{dy}{dx} \) on one side and solve for \( \frac{dy}{dx} \).

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Q3b. Find \( \frac{dy}{dx} \) for \( y = \left( \frac{x^2}{x^3 + 1} \right) \sqrt{x} \).

Background

Topic: Differentiation – Product and Quotient Rule

This question tests your ability to differentiate a function that is a product of a rational function and a root function.

Key Terms and Formulas:

• Product rule: \( (uv)' = u'v + uv' \)

• Quotient rule: \( \left( \frac{f}{g} \right)' = \frac{f'g - fg'}{g^2} \)

• Derivative of \( \sqrt{x} \): \( \frac{1}{2\sqrt{x}} \)

Step-by-Step Guidance

1. Let \( u = \frac{x^2}{x^3 + 1} \) and \( v = \sqrt{x} \).

2. Find \( u' \) using the quotient rule.

3. Find \( v' \) using the power rule.

4. Apply the product rule: \( y' = u'v + uv' \).

5. Simplify the expression as much as possible, but leave the final calculation for the student.

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Q3c. Find \( \frac{dy}{dx} \) for \( x^2 y^3 + \tan(x + y) = 1 \).

Background

Topic: Implicit Differentiation

This question tests your ability to use implicit differentiation for equations involving both x and y, including trigonometric functions.

Key Terms and Formulas:

• Implicit differentiation

• Chain rule

• Derivative of \( \tan(x + y) \): \( \sec^2(x + y) \cdot (1 + \frac{dy}{dx}) \)

Step-by-Step Guidance

1. Differentiate both sides of the equation with respect to x.

2. For \( x^2 y^3 \), use the product rule: \( 2x y^3 + x^2 \cdot 3y^2 \frac{dy}{dx} \).

3. For \( \tan(x + y) \), use the chain rule: \( \sec^2(x + y) (1 + \frac{dy}{dx}) \).

4. Set the derivative of 1 to zero.

5. Collect all \( \frac{dy}{dx} \) terms on one side and solve for \( \frac{dy}{dx} \).

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Q3d. Find \( \frac{dy}{dx} \) for \( \ln(xy) = e^{x^2 + y^2} \).

Background

Topic: Implicit Differentiation

This question tests your ability to use implicit differentiation for equations involving logarithmic and exponential functions.

Key Terms and Formulas:

• Derivative of \( \ln(xy) \): \( \frac{1}{xy} (y + x \frac{dy}{dx}) \)

• Derivative of \( e^{x^2 + y^2} \): \( e^{x^2 + y^2} (2x + 2y \frac{dy}{dx}) \)

Step-by-Step Guidance

1. Differentiate both sides of the equation with respect to x.

2. For \( \ln(xy) \), use the chain rule and product rule.

3. For \( e^{x^2 + y^2} \), use the chain rule.

4. Collect all \( \frac{dy}{dx} \) terms on one side and solve for \( \frac{dy}{dx} \).

5. Simplify the expression as much as possible, but leave the final calculation for the student.

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Q4a. For \( h(x) = \sqrt{f(x^2 + 1)} + 4 \), find \( h'(2) \) given values from the table.

Background

Topic: Chain Rule and Derivatives from a Table

This question tests your ability to use the chain rule and evaluate derivatives using values from a table.

Key Terms and Formulas:

• Chain rule: \( \frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x) \)

• Derivative of \( \sqrt{u} \): \( \frac{1}{2\sqrt{u}} \cdot u' \)

Step-by-Step Guidance

1. Let \( u = f(x^2 + 1) \), so \( h(x) = \sqrt{u} + 4 \).

2. Find \( h'(x) = \frac{1}{2\sqrt{u}} \cdot u' \).

3. Compute \( u' = f'(x^2 + 1) \cdot 2x \) using the chain rule.

4. Plug in \( x = 2 \) to find \( u \) and \( u' \) using the table values.

5. Combine the results to find \( h'(2) \).

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Q4b. For \( h(x) = g\left( \frac{x+1}{f(x)} \right) \), find \( h'(3) \) given values from the table.

Background

Topic: Chain Rule and Quotient Rule

This question tests your ability to use the chain rule and quotient rule, and evaluate derivatives using values from a table.

Key Terms and Formulas:

• Chain rule: \( h'(x) = g'(u) \cdot u' \), where \( u = \frac{x+1}{f(x)} \)

• Quotient rule: \( u' = \frac{f(x) \cdot 1 - (x+1) \cdot f'(x)}{[f(x)]^2} \)

Step-by-Step Guidance

1. Let \( u = \frac{x+1}{f(x)} \).

2. Find \( u' \) using the quotient rule.

3. Find \( h'(x) = g'(u) \cdot u' \).

4. Plug in \( x = 3 \) and use the table values for \( f(3) \), \( f'(3) \), and \( g'(u) \).

5. Combine the results to find \( h'(3) \).

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Q4c. For \( h(x) = (g(x^2 + x))^2 \), find \( h'(-2) \) given values from the table.

Background

Topic: Chain Rule and Power Rule

This question tests your ability to use the chain rule and power rule, and evaluate derivatives using values from a table.

Key Terms and Formulas:

• Chain rule: \( h'(x) = 2g(x^2 + x) \cdot g'(x^2 + x) \cdot (2x + 1) \)

Step-by-Step Guidance

1. Let \( u = x^2 + x \), so \( h(x) = (g(u))^2 \).

2. Find \( h'(x) = 2g(u) \cdot g'(u) \cdot u' \), where \( u' = 2x + 1 \).

3. Plug in \( x = -2 \) and use the table values for \( g(u) \) and \( g'(u) \).

4. Combine the results to find \( h'(-2) \).

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Q4d. For \( h(x) = |f(x) - 2| + \frac{(g(x))^2}{3} \), find \( h'(5) \) given values from the table.

Background

Topic: Derivatives and Absolute Value

This question tests your ability to differentiate functions involving absolute value and powers, using table values.

Key Terms and Formulas:

• Derivative of \( |u| \): \( \frac{u}{|u|} \cdot u' \) if \( u \neq 0 \)

• Power rule: \( \frac{d}{dx} [g(x)^2] = 2g(x)g'(x) \)

Step-by-Step Guidance

1. Let \( u = f(x) - 2 \), so \( \frac{d}{dx} |u| = \frac{u}{|u|} \cdot f'(x) \).

2. Find \( \frac{d}{dx} \frac{(g(x))^2}{3} = \frac{2g(x)g'(x)}{3} \).

3. Plug in \( x = 5 \) and use the table values for \( f(5) \), \( f'(5) \), \( g(5) \), and \( g'(5) \).

4. Combine the results to find \( h'(5) \).

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Q4e. For \( h(x) = \arccot(f(x)) - f(x)g'(x) \), find \( h'(1) \) given values from the table.

Background

Topic: Chain Rule and Derivatives of Inverse Trig Functions

This question tests your ability to differentiate functions involving inverse trigonometric functions and products, using table values.

Key Terms and Formulas:

• Derivative of \( \arccot(u) \): \( -\frac{1}{1 + u^2} \cdot u' \)

• Product rule: \( \frac{d}{dx} [f(x)g'(x)] = f'(x)g'(x) + f(x)g''(x) \)

Step-by-Step Guidance

1. Let \( u = f(x) \), so \( \frac{d}{dx} \arccot(u) = -\frac{1}{1 + u^2} \cdot f'(x) \).

2. Differentiate \( f(x)g'(x) \) using the product rule.

3. Plug in \( x = 1 \) and use the table values for \( f(1) \), \( f'(1) \), \( g'(1) \), and \( g''(1) \).

4. Combine the results to find \( h'(1) \).

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Q5. Let \( f(x) = \frac{3x+1}{2x-1} \). Find \( (f^{-1})'(2) \).

Background

Topic: Derivative of Inverse Functions

This question tests your ability to find the derivative of the inverse function at a given point.

Key Terms and Formulas:

• Derivative of inverse: \( (f^{-1})'(a) = \frac{1}{f'(f^{-1}(a))} \)

Step-by-Step Guidance

1. Find the value \( x \) such that \( f(x) = 2 \).

2. Compute \( f'(x) \) using the quotient rule.

3. Plug the value of \( x \) into \( f'(x) \).

4. Use the formula \( (f^{-1})'(2) = \frac{1}{f'(x)} \).

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Q6. Let \( g(x) = f^{-1}(x) \). Suppose \( f(3) = 4, f'(3) = 5, f(4) = 6, f'(4) = 7 \). Find the derivative of \( h(x) = (g(x))^2 \) evaluated at \( x = 4 \).

Background

Topic: Derivative of Inverse Functions and Chain Rule

This question tests your ability to use the chain rule and properties of inverse functions to find the derivative at a specific value.

Key Terms and Formulas:

• Chain rule: \( h'(x) = 2g(x)g'(x) \)

• Derivative of inverse: \( g'(x) = \frac{1}{f'(g(x))} \)

Step-by-Step Guidance

1. Find \( g(4) \), which is the value \( a \) such that \( f(a) = 4 \).

2. Find \( f'(g(4)) \) using the given values.

3. Compute \( h'(4) = 2g(4)g'(4) \).

4. Plug in the values for \( g(4) \) and \( g'(4) \).

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Q7. Given \( y = \frac{(3x+2)^4}{(x-1)^2 \sqrt{4x+5}} \), find \( \frac{dy}{dx} \) using implicit differentiation. Leave your answer in terms of x.

Background

Topic: Differentiation – Quotient Rule and Chain Rule

This question tests your ability to differentiate a function involving a quotient and a root, using the quotient rule and chain rule.

Key Terms and Formulas:

• Quotient rule: \( \left( \frac{f}{g} \right)' = \frac{f'g - fg'}{g^2} \)

• Chain rule for powers and roots

Step-by-Step Guidance

1. Let \( f(x) = (3x+2)^4 \) and \( g(x) = (x-1)^2 \sqrt{4x+5} \).

2. Find \( f'(x) \) using the chain rule.

3. Find \( g'(x) \) using the product rule and chain rule.

4. Apply the quotient rule: \( y' = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \).

5. Simplify the expression as much as possible, but leave the final calculation for the student.

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Q8A. Find the general antiderivative of \( f(x) = \sec^2(x) e^{\tan(x)} \).

Background

Topic: Integration – Substitution

This question tests your ability to find the antiderivative using substitution for a function involving trigonometric and exponential terms.

Key Terms and Formulas:

• Substitution: Let \( u = \tan(x) \), then \( du = \sec^2(x) dx \).

• Integral of \( e^u du = e^u + C \)

Step-by-Step Guidance

1. Let \( u = \tan(x) \), so \( du = \sec^2(x) dx \).

2. Rewrite the integral in terms of u: \( \int \sec^2(x) e^{\tan(x)} dx = \int e^u du \).

3. Integrate \( e^u \) with respect to u.

4. Substitute back \( u = \tan(x) \) after integrating.

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Q8B. Find the general antiderivative of \( f(x) = x \cdot 5^{x^2} \).

Background

Topic: Integration – Substitution and Exponential Functions

This question tests your ability to find the antiderivative using substitution for a function involving an exponential with a quadratic exponent.

Key Terms and Formulas:

• Substitution: Let \( u = x^2 \), then \( du = 2x dx \).

• Integral of \( a^u du = \frac{a^u}{\ln a} + C \)

Step-by-Step Guidance

1. Let \( u = x^2 \), so \( du = 2x dx \).

2. Rewrite the integral: \( \int x \cdot 5^{x^2} dx = \frac{1}{2} \int 5^u du \).

3. Integrate \( 5^u \) with respect to u.

4. Substitute back \( u = x^2 \) after integrating.

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Q8C. Find the general antiderivative of \( f(x) = \frac{x^2}{(1 - 2x^3)^2} \).

Background

Topic: Integration – Substitution

This question tests your ability to find the antiderivative using substitution for a rational function with a cubic denominator.

Key Terms and Formulas:

• Substitution: Let \( u = 1 - 2x^3 \), then \( du = -6x^2 dx \).

• Integral of \( u^{-2} du = -u^{-1} + C \)

Step-by-Step Guidance

1. Let \( u = 1 - 2x^3 \), so \( du = -6x^2 dx \).

2. Rewrite the integral: \( \int \frac{x^2}{(1 - 2x^3)^2} dx = -\frac{1}{6} \int u^{-2} du \).

3. Integrate \( u^{-2} \) with respect to u.

4. Substitute back \( u = 1 - 2x^3 \) after integrating.

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Q8D. Find the general antiderivative of \( f(x) = \frac{e^{2x}}{e^{2x} - 1} \).

Background

Topic: Integration – Substitution and Rational Functions

This question tests your ability to find the antiderivative using substitution for a rational function involving exponentials.

Key Terms and Formulas:

• Substitution: Let \( u = e^{2x} - 1 \), then \( du = 2e^{2x} dx \).

• Integral of \( \frac{1}{u} du = \ln|u| + C \)

Step-by-Step Guidance

1. Let \( u = e^{2x} - 1 \), so \( du = 2e^{2x} dx \).

2. Rewrite the integral: \( \int \frac{e^{2x}}{e^{2x} - 1} dx = \frac{1}{2} \int \frac{1}{u} du \).

3. Integrate \( \frac{1}{u} \) with respect to u.

4. Substitute back \( u = e^{2x} - 1 \) after integrating.

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