# 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 $$x=a$$.
*** 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: * $$f(x)=x^2 \Rightarrow f'(x)=2x^{1}=2x$$ * $$f(x)=\sqrt{x}=x^{1/2} \Rightarrow f'(x)=\frac{1}{2}x^{-1/2}=\frac{1}{2\sqrt{x}}$$ * $$f(x)=1/x = x^{-1} \Rightarrow f'(x)=-1 \cdot x^{-2}=-1/x^2$$
*** 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: $$c = c \cdot x^0 \Rightarrow \frac{d}{dx}(c \cdot x^0) = c \cdot 0 \cdot x^{-1} = 0$$.
*** 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: $$\frac{d}{dx}(a^x) = a^x \ln a$$.
*** 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: $$\frac{d}{dx}(\log\_a x) = \frac{1}{x \ln a}$$.
*** ### 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)$$. * Differentiation is a linear operation, meaning the derivative of a sum is the sum of the derivatives.
*** 7\. The Product Rule for differentiation states: $$\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'$$ gives the same result.
*** 8\. The Quotient Rule for $$\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}$$. * A common mnemonic is "Low d-High minus High d-Low, over the square of what's below." * Be careful with the order and the sign in the numerator. It is "derivative of the top times the bottom, minus the top times derivative of the bottom."
*** ### Chain Rule 9\. The Chain Rule for the derivative of a composite function $$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$$.
*** 10\. Using the Chain Rule, the derivative of $$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$$.
*** ### Higher-Order Derivatives 11\. The second derivative of a function $$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$$, the graph is concave down (like a cap).
*** ### Implicit Differentiation 12\. Implicit differentiation is used when: 1. The function is given explicitly as $$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}$$. * This technique is essential for finding derivatives of inverse functions and for curves defined implicitly.
*** 13\. Using implicit differentiation on $$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}$$
*** ### Introduction to Partial Derivatives 14\. For a function of two variables, $$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}$$ is found by treating x as a constant and differentiating with respect to y.
*** 15\. For $$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$$.
*** ### Higher-Order Partial Derivatives 16\. For a function $$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}$$.