Basic Derivative Rules - Calculus 1

Let f and h be differentiable functions. Let k(x) = 3h(x) - f(4). Given h'(x) = f(x), f(4) = 10, and f'(4) = -2. What is k'(4)?
Answer: 30 First, find the derivative of k(x). Since f(4) is a constant value, its derivative is 0. Using the Constant Multiple and Difference rules, k'(x) = 3h'(x) - 0 = 3h'(x). We are given that h'(x) = f(x), so we can substitute to get k'(x) = 3f(x). To find k'(4), we evaluate this expression at x=4: k'(4) = 3f(4) = 3(10) = 30.
Let g(t) = A·sin(t) - B, where A = cos(k) and B = ln(k²+1) for some constant k. Which of the following is g'(t)? Assume that the derivative of sin(t) with respect to t is cos(t).
Answer: cos(k)cos(t) The function is g(t) and we are differentiating with respect to t. The expressions A=cos(k) and B=ln(k²+1) do not depend on t, so they are treated as constants. Applying the derivative rules: g'(t) = d/dt [A·sin(t) - B] = A·(d/dt[sin(t)]) - d/dt[B] = A·cos(t) - 0. Substituting A = cos(k) gives g'(t) = cos(k)cos(t).
Let f(x) be a differentiable function and let C be a non-zero constant. If h(x) = f(x) + C, what is the relationship between the slope of the tangent line to f(x) at x=a and the slope of the tangent line to h(x) at x=a?
Answer: The slopes are identical. The slope of the tangent line at x=a is given by the derivative at that point. The slope for f(x) is f'(a). For h(x), the slope is h'(a). We find the derivative h'(x) using the Sum and Constant rules: h'(x) = f'(x) + 0 = f'(x). Therefore, h'(a) = f'(a) for any value of a. The slopes are identical. Geometrically, adding a constant C to a function shifts its graph vertically, which does not change the slope at any point.
Let $f(x) = x^3 - 5x^2 + x + 8$ . What is the value of $f'(1) + f''(1)$ ?
Answer: $-10$ First, we find the first and second derivatives of $f(x)$ . $f'(x) = 3x^2 - 10x + 1$ $f''(x) = 6x - 10$ Next, we evaluate both derivatives at $x=1$ . $f'(1) = 3(1)^2 - 10(1) + 1 = 3 - 10 + 1 = -6$ $f''(1) = 6(1) - 10 = 6 - 10 = -4$ Finally, we add the two values: $f'(1) + f''(1) = -6 + (-4) = -10$
Let f(x) be a function whose derivative is g(x), i.e., f'(x)=g(x). Let h(x) be a function whose derivative is k(x), i.e., h'(x)=k(x). If p(x) = 3f(x) - 5h(x), what is p'(x)?
Answer: 3g(x) - 5k(x) This question tests your understanding of differentiation rules, specifically the constant multiple rule and the sum/difference rule. When you see a function that's a linear combination of other functions, you need to apply these rules systematically. To find $p'(x)$ , you differentiate $p(x) = 3f(x) - 5h(x)$ term by term. Using the constant multiple rule, the derivative of $3f(x)$ is $3 \cdot f'(x)$ , and the derivative of $5h(x)$ is $5 \cdot h'(x)$ . Since we're given that $f'(x) = g(x)$ and $h'(x) = k(x)$ , we can substitute: $p'(x) = 3f'(x) - 5h'(x) = 3g(x) - 5k(x)$ . Looking at the wrong answers: Choice A ( $g'(x) - k'(x)$ ) incorrectly takes the derivatives of the derivatives, going one step too far. Choice B ( $g(x) - k(x)$ ) ignores the constant multipliers 3 and 5 entirely, forgetting the constant multiple rule. Choice C ( $3f'(x) - 5h'(x)$ ) is actually correct mathematically but doesn't use the given substitutions $f'(x) = g(x)$ and $h'(x) = k(x)$ , so it's not simplified to match the answer format. Remember this pattern: when differentiating linear combinations, constants stay put and multiply the derivatives. Always substitute any given relationships (like $f'(x) = g(x)$ ) to express your final answer in the requested form. Practice recognizing when the question wants you to use specific substitutions rather than leaving derivatives in their original notation.
The derivative of a function $f(x)$ is given by $f'(x) = 6x^2 - 4$ . If the graph of $f(x)$ passes through the point $(1, 5)$ , which of the following is the expression for $f(x)$ ?
Answer: $f(x) = 2x^3 - 4x + 7$ We need to find a function $f(x)$ whose derivative is $f'(x) = 6x^2 - 4$ . By reversing the power rule, the general form of $f(x)$ is $2x^3 - 4x + C$ for some constant $C$ . We are given that $f(1) = 5$ . We can use this to solve for $C$ : $f(1) = 2(1)^3 - 4(1) + C = 5$ $2 - 4 + C = 5$ $-2 + C = 5$ $C = 7$ Therefore, the function is $f(x) = 2x^3 - 4x + 7$ .
Let f(x) = 4sin(x) + 3eˣ. What is the slope of the tangent line to the graph of f at x=0? Assume that the derivative of sin(x) is cos(x) and the derivative of eˣ is eˣ.
Answer: 7 The slope of the tangent line at x=0 is given by f'(0). First, find the derivative function f'(x) using the Sum and Constant Multiple rules: f'(x) = 4cos(x) + 3eˣ. Then, evaluate f'(x) at x=0: f'(0) = 4cos(0) + 3e⁰ = 4(1) + 3(1) = 7.
Let u(x) and v(x) be differentiable functions of x. If y = 5u(x) - 2v(x) + e, where e is the base of the natural logarithm, what is dy/dx?
Answer: 5(du/dx) - 2(dv/dx) We are asked to find the derivative of y with respect to x. Using the Sum, Difference, and Constant Multiple rules, we differentiate term by term: dy/dx = d/dx[5u(x)] - d/dx[2v(x)] + d/dx[e]. This gives dy/dx = 5(du/dx) - 2(dv/dx) + 0, because e is a constant. The result is dy/dx = 5(du/dx) - 2(dv/dx).
Let $p(x) = -2x^4 + x^3 - 5x^2 + 10x - 3$ . What is the least positive integer $n$ for which the $n$ -th derivative, $p^{(n)}(x)$ , is equal to zero for all $x$ ?
Answer: 5 We find successive derivatives of the 4th-degree polynomial $p(x)$ . $p'(x) = -8x^3 + 3x^2 - 10x + 10$ $p''(x) = -24x^2 + 6x - 10$ $p'''(x) = -48x + 6$ $p^{(4)}(x) = -48$ $p^{(5)}(x) = 0$ The fifth derivative is the first derivative that is equal to zero for all $x$ . Therefore, the least positive integer is $n=5$ . In general, for a polynomial of degree $k$ , the $(k+1)$ -th derivative is zero.
A function is defined by the polynomial $p(x) = ax^3 + bx^2 + cx + d$ . The function satisfies the conditions $p(0) = 5$ , $p'(0) = -3$ , and $p''(0) = 8$ . What is the value of the coefficient $b$ ?
Answer: 4 We find the first and second derivatives of the general polynomial $p(x)$ . $p'(x) = 3ax^2 + 2bx + c$ $p''(x) = 6ax + 2b$ Now we use the given conditions. The condition $p''(0)=8$ allows us to find $b$ . $p''(0) = 6a(0) + 2b = 8$ $2b = 8$ $b = 4$ The other conditions determine the other coefficients: $p(0) = d = 5$ and $p'(0) = c = -3$ . The question only asks for the value of $b$ .
Let f(x) and g(x) be differentiable functions with h'(x) = 5f'(x) - 2g'(x). If h(0) = 1, f(0) = 2, and g(0) = 3, which of the following could be an expression for h(x)?
Answer: 5f(x) - 2g(x) - 3 This problem tests your understanding of the relationship between derivatives and antiderivatives, specifically how differentiation rules work in reverse. Given that $h'(x) = 5f'(x) - 2g'(x)$ , you need to find which expression for $h(x)$ has this derivative. Using the linearity of differentiation, if $h(x) = 5f(x) - 2g(x) + C$ (where $C$ is a constant), then $h'(x) = 5f'(x) - 2g'(x)$ since the derivative of a constant is zero. This confirms the general form must be $5f(x) - 2g(x) + C$ . To find the specific value of $C$ , use the initial conditions. You know $h(0) = 1,f(0) = 2$ , and $g(0) = 3$ . Substituting into $h(x) = 5f(x) - 2g(x) + C$ : $1 = 5(2) - 2(3) + C$ $1 = 10 - 6 + C$ $1 = 4 + C$ $C = -3$ Therefore, $h(x) = 5f(x) - 2g(x) - 3$ , which is answer choice D. Choice A lacks the constant term needed to satisfy the initial condition. Choice B gives $h(0) = 5(2) - 2(3) + 3 = 7 \neq 1$ . Choice C gives $h(0) = 5(2) - 2(3) - 4 = 0 \neq 1$ . Study tip: When working backwards from a derivative to find the original function, remember that antiderivatives always include a constant of integration. Use initial conditions to determine this constant's specific value.
Let f(x) be a differentiable function. If g(x) = f(x) + f(a), where a is a constant, which of the following is g'(a)?
Answer: f'(a) When you encounter a function defined as the sum of terms involving another function, you need to carefully apply differentiation rules to each component. Given $g(x) = f(x) + f(a)$ , let's find $g'(a)$ by first finding $g'(x)$ for any value of $x$ , then evaluating at $x = a$ . To find $g'(x)$ , differentiate each term separately. The derivative of $f(x)$ with respect to $x$ is simply $f'(x)$ . However, $f(a)$ is a constant because $a$ is a constant—it doesn't depend on $x$ at all. The derivative of any constant is zero. Therefore: $g'(x) = f'(x) + 0 = f'(x)$ Now evaluating at $x = a$ : $g'(a) = f'(a)$ Looking at the incorrect options: Choice A, $f'(a) + f(a)$ , mistakenly treats $f(a)$ as if it contributes $f(a)$ to the derivative rather than zero. This confuses the function value with its derivative. Choice B, $2f'(a)$ , incorrectly assumes both terms contribute $f'(a)$ to the derivative, forgetting that $f(a)$ is constant. Choice D suggests the entire derivative is zero, which would only be true if $g(x)$ were entirely constant. The correct answer is C: $f'(a)$ . Study tip: Remember that when differentiating, any term that doesn't contain the variable of differentiation is treated as a constant, and constants always differentiate to zero. Focus on identifying which parts of a function actually depend on your variable.
Let f and g be differentiable functions with f'(1) = -2 and g'(1) = 3. If h(x) = 4g(x) - 2f(x) + π², what is the value of h'(1)?
Answer: 16 Using the Sum, Difference, Constant Multiple, and Constant rules, we find the derivative of h(x): h'(x) = 4g'(x) - 2f'(x) + 0. Note that π² is a constant, so its derivative is zero. Evaluating at x=1 gives: h'(1) = 4g'(1) - 2f'(1) = 4(3) - 2(-2) = 12 + 4 = 16.
Let $f(x)$ be a differentiable function and let $A$ and $B$ be constants. A function $h(x)$ is defined as $h(x) = A \cdot f(x) + B$ . Given that $h(2) = 10$ , $h'(2) = 12$ , $f(2) = 1$ , and $f'(2) = 3$ , what is the value of the constant $A$ ?
Answer: 4 The question asks for the value of the constant $A$ . We need to use the information about the derivatives. First, find the derivative of $h(x)$ using the constant multiple and constant rules: $h'(x) = A \cdot f'(x) + 0 = A \cdot f'(x)$ . Now, substitute the given values at $x=2$ : $h'(2) = A \cdot f'(2)$ . We are given $h'(2)=12$ and $f'(2)=3$ . So, the equation becomes $12 = A \cdot 3$ . Solving for $A$ , we get $A = 12 / 3 = 4$ . The information $h(2)=10$ and $f(2)=1$ is extra information not needed to find $A$ .
Let $k$ be a non-zero real constant. If $g(t) = 2t^k - kt^4 + k^3$ , what is the derivative $g'(t)$ ?
Answer: $2kt^{k-1} - 4kt^3$ We differentiate $g(t)$ with respect to $t$ , treating $k$ as a constant. For the first term, $2t^k$ , we use the power rule. For the second term, $-kt^4$ , we use the constant multiple and power rules. For the third term, $k^3$ , since $k$ is a constant, $k^3$ is also a constant, and its derivative is 0. $g'(t) = \frac{d}{dt}(2t^k) - \frac{d}{dt}(kt^4) + \frac{d}{dt}(k^3)$ $g'(t) = 2(kt^{k-1}) - k(4t^3) + 0$ $g'(t) = 2kt^{k-1} - 4kt^3$
Let D(f) denote the derivative of a function f. Let h(x) = 3f(x) - 2g(x). Which of the following is equivalent to D(D(h)) evaluated at x=a?
Answer: 3f''(a) - 2g''(a) D(h) is the first derivative of h, which is h'(x). Using the basic derivative rules, h'(x) = 3f'(x) - 2g'(x). D(D(h)) means taking the derivative of D(h) = h'(x), which gives the second derivative h''(x). Differentiating h'(x), we get h''(x) = 3f''(x) - 2g''(x). Evaluating this expression at x=a gives 3f''(a) - 2g''(a).
Let D be the differentiation operator, such that D(f) = f'. Which of the following statements best describes a property of this operator for any differentiable functions f and g and any constants a and b?
Answer: D(a·f + b·g) = a·D(f) + b·D(g) This question asks to identify the linearity property of differentiation. The Sum Rule states D(f+g) = D(f) + D(g), and the Constant Multiple Rule states D(c·f) = c·D(f). Combining these, we get D(a·f + b·g) = D(a·f) + D(b·g) = a·D(f) + b·D(g). This property is known as linearity. Option D is the product rule, which is a different property.
Let k be a real number. If f(x) = k - eˣ, what is f'(ln k)? Assume the derivative of eˣ is eˣ.
Answer: -k This problem tests your ability to find derivatives of exponential functions and then evaluate them at specific points. To find $f'(x)$ , you need to differentiate $f(x) = k - e^x$ . Since $k$ is a constant, its derivative is zero. The derivative of $e^x$ is $e^x$ , so $f'(x) = 0 - e^x = -e^x$ . Now you need to evaluate $f'(\ln k)$ . Substituting $x = \ln k$ into your derivative: $f'(\ln k) = -e^{\ln k}$ . Here's the key insight: $e^{\ln k} = k$ because exponential and natural logarithm are inverse functions. Therefore, $f'(\ln k) = -k$ . Looking at the wrong answers: Choice (A) gives $k$ , which would be correct if you forgot the negative sign from differentiating $-e^x$ . Choice (B) gives $1-k$ , which might result from incorrectly thinking the derivative of $k$ is $1$ instead of $0$ , then computing $1 - k$ . Choice (C) gives $0$ , which could happen if you mistakenly thought you were evaluating $f(\ln k)$ instead of $f'(\ln k)$ and tried to use some property incorrectly. Remember that $e^{\ln a} = a$ for any positive real number $a$ . This fundamental relationship between exponential and logarithmic functions appears frequently in calculus problems, so make sure you can recognize and apply it quickly when evaluating expressions.
Let h(θ) = 2sin(θ) + 4cos(θ). Find the value of h'(π/6). Assume the derivative of sin(θ) is cos(θ) and the derivative of cos(θ) is -sin(θ).
Answer: √3 - 2 First, find the derivative of h(θ): h'(θ) = 2cos(θ) + 4(-sin(θ)) = 2cos(θ) - 4sin(θ). Next, substitute θ = π/6 into h'(θ): h'(π/6) = 2cos(π/6) - 4sin(π/6). Using the values cos(π/6) = √3/2 and sin(π/6) = 1/2, we get: h'(π/6) = 2(√3/2) - 4(1/2) = √3 - 2.
Let $f(x)$ and $g(x)$ be differentiable functions. Two new functions are defined as $h_1(x) = 2f(x) + 3g(x)$ and $h_2(x) = 3f(x) + 2g(x)$ . Suppose that for some constant $c$ , we have $f'(c) > g'(c) > 0$ . Which of the following statements must be true?
Answer: $h_2'(c) > h_1'(c)$ When you see functions combined linearly like this, you're dealing with differentiation rules for linear combinations. The key insight is recognizing how coefficients affect the derivatives when you have competing rates of change. First, let's find the derivatives using the linearity of differentiation. For $h_1(x) = 2f(x) + 3g(x)$ , we get $h_1'(x) = 2f'(x) + 3g'(x)$ . For $h_2(x) = 3f(x) + 2g(x)$ , we get $h_2'(x) = 3f'(x) + 2g'(x)$ . At point $c$ , this gives us $h_1'(c) = 2f'(c) + 3g'(c)$ and $h_2'(c) = 3f'(c) + 2g'(c)$ . Since $f'(c) > g'(c)$ , we can compare these expressions by looking at their difference: $h_2'(c) - h_1'(c) = [3f'(c) + 2g'(c)] - [2f'(c) + 3g'(c)] = f'(c) - g'(c)$ . Because $f'(c) > g'(c)$ , we have $f'(c) - g'(c) > 0$ , which means $h_2'(c) > h_1'(c)$ . Choice A is correct because $h_2$ gives more weight (coefficient 3) to the faster-growing function $f$ . Choice B reverses this relationship incorrectly. Choice C would only be true if $f'(c) = g'(c)$ , which contradicts our given condition. Choice D is wrong because we have sufficient information to determine the relationship definitively. Study tip: When comparing linear combinations of functions, look at which combination gives more weight to the terms with larger derivatives—that combination will have the larger overall derivative.

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