4.2 MCQs-Ordinary & Partial Differentiation

Ordinary & Partial Differentiation MCQs

Basic Rules of Differentiation

1. The derivative of a function $f(x)$ at a point $x = a$, denoted $f'(a)$, is defined as:

1. $f(a)$

2. $\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$

3. $\frac{f(a+h) - f(a)}{h}$

4. $f(a+h) - f(a)$

Show me the answer

Answer: 2. $\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$

Explanation:

• This is the formal definition of the derivative as the limit of the difference quotient.

• It represents the instantaneous rate of change of $f$ at $x=a$, or the slope of the tangent line to the curve $y=f(x)$ at that point.

• If this limit exists, the function is said to be differentiable at $x=a$.

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2. If $f(x) = x^n$, where n is a constant, then $f'(x)$ is:

1. $nx$

2. $nx^{n-1}$

3. $(n-1)x^n$

4. $x^{n-1}$

Show me the answer

Answer: 2. $nx^{n-1}$

Explanation:

• This is the Power Rule, one of the most fundamental differentiation rules.

• It applies to any real number exponent n.

• Examples:

  • $f(x)=x^2 \Rightarrow f'(x)=2x^{1}=2x$

  • $f(x)=\sqrt{x}=x^{1/2} \Rightarrow f'(x)=\frac{1}{2}x^{-1/2}=\frac{1}{2\sqrt{x}}$

  • $f(x)=1/x = x^{-1} \Rightarrow f'(x)=-1 \cdot x^{-2}=-1/x^2$

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3. The derivative of a constant function $f(x) = c$ is:

1. $c$

2. $x$

3. 0

4. 1

Show me the answer

Answer: 3. 0

Explanation:

• The graph of a constant function is a horizontal line. The slope of a horizontal line is 0.

• Formally, using the limit definition: $f'(x) = \lim_{h \to 0} \frac{c - c}{h} = \lim_{h \to 0} 0 = 0$.

• This aligns with the Power Rule: $c = c \cdot x^0 \Rightarrow \frac{d}{dx}(c \cdot x^0) = c \cdot 0 \cdot x^{-1} = 0$.

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4. The derivative of $f(x) = e^x$ is:

1. $x e^{x-1}$

2. $e^x$

3. $\ln x$

4. $0$

Show me the answer

Answer: 2. $e^x$

Explanation:

• The exponential function $f(x) = e^x$ is unique because it is its own derivative: $\frac{d}{dx}(e^x) = e^x$.

• This property makes it extremely important in calculus, differential equations, and modeling growth/decay.

• In contrast, for a general exponential: $\frac{d}{dx}(a^x) = a^x \ln a$.

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5. The derivative of $f(x) = \ln x$ (for $x > 0$) is:

1. $e^x$

2. $\frac{1}{x}$

3. $x$

4. $\frac{1}{\ln x}$

Show me the answer

Answer: 2. $\frac{1}{x}$

Explanation:

• The natural logarithm function has a simple derivative: $\frac{d}{dx}(\ln x) = \frac{1}{x}$.

• This result is derived using implicit differentiation or the definition of $\ln x$ as the inverse of $e^x$.

• For a general logarithm: $\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}$.

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Rules for Combinations of Functions

6. If $f(x)$ and $g(x)$ are differentiable functions, the derivative of their sum is given by:

1. $f'(x) + g'(x)$

2. $f'(x) g'(x)$

3. $f(x) g'(x) + f'(x) g(x)$

4. $\frac{f'(x) g(x) - f(x) g'(x)}{[g(x)]^2}$

Show me the answer

Answer: 1. $f'(x) + g'(x)$

Explanation:

• This is the Sum Rule: $\frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)$.

• Similarly, the Difference Rule is: $\frac{d}{dx}[f(x) - g(x)] = f'(x) - g'(x)$.

• Differentiation is a linear operation, meaning the derivative of a sum is the sum of the derivatives.

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7. The Product Rule for differentiation states: $\frac{d}{dx}[f(x) g(x)] =$

1. $f'(x) g'(x)$

2. $f(x) g'(x)$

3. $f'(x) g(x) + f(x) g'(x)$

4. $f'(x) g(x) - f(x) g'(x)$

Show me the answer

Answer: 3. $f'(x) g(x) + f(x) g'(x)$

Explanation:

• The Product Rule is: $(fg)' = f'g + fg'$.

• In words: The derivative of a product is "derivative of the first times the second, plus the first times the derivative of the second."

• Order matters in the sense of which function is "first," but the rule is symmetric: $(gf)' = g'f + gf'$ gives the same result.

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8. The Quotient Rule for $\frac{f(x)}{g(x)}$ (where $g(x) \neq 0$) is:

1. $\frac{f'(x)}{g'(x)}$

2. $\frac{f'(x) g(x) - f(x) g'(x)}{[g(x)]^2}$

3. $\frac{f(x) g'(x) - f'(x) g(x)}{[g(x)]^2}$

4. $f'(x) g(x) - f(x) g'(x)$

Show me the answer

Answer: 2. $\frac{f'(x) g(x) - f(x) g'(x)}{[g(x)]^2}$

Explanation:

• The Quotient Rule formula is: $\left( \frac{f}{g} \right)' = \frac{f'g - fg'}{g^2}$.

• A common mnemonic is "Low d-High minus High d-Low, over the square of what's below."

• Be careful with the order and the sign in the numerator. It is "derivative of the top times the bottom, minus the top times derivative of the bottom."

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Chain Rule

9. The Chain Rule for the derivative of a composite function $y = f(g(x))$ is:

1. $f'(g(x))$

2. $f'(x) \cdot g'(x)$

3. $f'(g(x)) \cdot g'(x)$

4. $f(g'(x))$

Show me the answer

Answer: 3. $f'(g(x)) \cdot g'(x)$

Explanation:

• If $y = f(u)$ and $u = g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = f'(u) \cdot g'(x) = f'(g(x)) \cdot g'(x)$.

• In Leibniz notation, it looks intuitive: $\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$.

• Example: Find $\frac{d}{dx} \sin(x^2)$. Let $u = x^2$, then $\frac{d}{dx} \sin(u) = \cos(u) \cdot \frac{du}{dx} = \cos(x^2) \cdot 2x$.

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10. Using the Chain Rule, the derivative of $y = (3x^2 + 5)^4$ with respect to x is:

1. $4(3x^2+5)^3$

2. $4(6x)^3$

3. $4(3x^2+5)^3 \cdot 6x$

4. $(3x^2+5)^3 \cdot 6x$

Show me the answer

Answer: 3. $4(3x^2+5)^3 \cdot 6x$

Explanation:

• Identify the outer function: $f(u) = u^4$, where $u = 3x^2 + 5$.

• Outer derivative: $f'(u) = 4u^3$.

• Inner derivative: $\frac{du}{dx} = 6x$.

• Apply Chain Rule: $\frac{dy}{dx} = f'(u) \cdot \frac{du}{dx} = 4(3x^2+5)^3 \cdot 6x$.

• This simplifies to $24x(3x^2+5)^3$.

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Higher-Order Derivatives

11. The second derivative of a function $f(x)$, denoted $f''(x)$ or $\frac{d^2y}{dx^2}$, represents:

1. The slope of the tangent line

2. The rate of change of the function's derivative

3. The area under the curve

4. The y-intercept

Show me the answer

Answer: 2. The rate of change of the function's derivative

Explanation:

• The first derivative $f'(x)$ represents velocity (rate of change of position).

• The second derivative $f''(x)$ represents acceleration (rate of change of velocity).

• Geometrically, the second derivative tells us about the concavity of the graph of $f$.

  • If $f''(x) > 0$, the graph is concave up (like a cup).

  • If $f''(x) < 0$, the graph is concave down (like a cap).

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Implicit Differentiation

12. Implicit differentiation is used when:

1. The function is given explicitly as $y = f(x)$

2. The relationship between x and y is given by an equation not solved for y

3. Finding the derivative of a constant function

4. Applying the Power Rule

Show me the answer

Answer: 2. The relationship between x and y is given by an equation not solved for y

Explanation:

• In implicit differentiation, we differentiate both sides of an equation like $x^2 + y^2 = 25$ with respect to x.

• When differentiating a term involving y (like $y^2$), we use the Chain Rule: $\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}$.

• We then solve the resulting equation for $\frac{dy}{dx}$.

• This technique is essential for finding derivatives of inverse functions and for curves defined implicitly.

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13. Using implicit differentiation on $x^2 + y^2 = 25$, the expression for $\frac{dy}{dx}$ is:

1. $\frac{x}{y}$

2. $-\frac{x}{y}$

3. $\frac{y}{x}$

4. $2x + 2y$

Show me the answer

Answer: 2. $-\frac{x}{y}$

Explanation:

• Differentiate both sides with respect to x: $\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)$

• Apply rules: $2x + 2y \frac{dy}{dx} = 0$

• Solve for $\frac{dy}{dx}$: $2y \frac{dy}{dx} = -2x$ $\frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y}$

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Introduction to Partial Derivatives

14. For a function of two variables, $z = f(x, y)$, the partial derivative $\frac{\partial f}{\partial x}$ is found by:

1. Treating y as a constant and differentiating with respect to x

2. Treating x as a constant and differentiating with respect to y

3. Differentiating with respect to x and then y

4. Adding the derivatives with respect to x and y

Show me the answer

Answer: 1. Treating y as a constant and differentiating with respect to x

Explanation:

• The partial derivative $f_x = \frac{\partial f}{\partial x}$ measures the rate of change of f as x changes, while holding y fixed.

• Geometrically, it gives the slope of the tangent line to the curve formed by intersecting the surface $z=f(x,y)$ with a plane $y = \text{constant}$.

• Similarly, $f_y = \frac{\partial f}{\partial y}$ is found by treating x as a constant and differentiating with respect to y.

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15. For $f(x, y) = 3x^2y + xy^3$, the partial derivative $f_x$ is:

1. $6xy + y^3$

2. $3x^2 + 3xy^2$

3. $6xy + 3xy^2$

4. $6x + 3y^2$

Show me the answer

Answer: 1. $6xy + y^3$

Explanation:

• To find $f_x$, treat y as a constant.

• $\frac{\partial}{\partial x}(3x^2y) = 3y \cdot \frac{d}{dx}(x^2) = 3y \cdot 2x = 6xy$

• $\frac{\partial}{\partial x}(xy^3) = y^3 \cdot \frac{d}{dx}(x) = y^3 \cdot 1 = y^3$

• Therefore, $f_x = 6xy + y^3$.

• Note: $f_y = 3x^2 + 3xy^2$.

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Higher-Order Partial Derivatives

16. For a function $f(x, y)$, the mixed second-order partial derivative $f_{xy}$ means:

1. Differentiate twice with respect to x

2. Differentiate twice with respect to y

3. Differentiate first with respect to x, then with respect to y

4. Differentiate first with respect to y, then with respect to x

Show me the answer

Answer: 3. Differentiate first with respect to x, then with respect to y

Explanation:

• Notation: $f_{xy} = \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial x} \right) = \frac{\partial^2 f}{\partial y \partial x}$.

• The order of differentiation in the notation is from right to left: $\frac{\partial^2 f}{\partial y \partial x}$ means "first differentiate with respect to x, then differentiate the result with respect to y."

• Under mild continuity conditions (Clairaut's Theorem), the mixed partials are equal: $f_{xy} = f_{yx}$.