Selecting Derivative Procedures - Calculus 1

1. Let $h(x) = f(g(x))$, where $g(2) = 3$, $g'(2) = -1$, $f'(2) = 4$, and $f'(3) = 5$. Which calculation correctly finds $h'(2)$?
   Answer: $h'(2) = f'(g(2)) \cdot g'(2) = f'(3) \cdot (-1)$ This is a direct application of the chain rule. The formula is $h'(x) = f'(g(x)) \cdot g'(x)$. To find $h'(2)$, we substitute $x=2$: $h'(2) = f'(g(2)) \cdot g'(2)$. Using the given values, $g(2)=3$ and $g'(2)=-1$, so $h'(2) = f'(3) \cdot (-1)$. Since $f'(3)=5$, the result is -5. Choice A shows the correct procedure.
2. If $x \tan(y) = y^2 + 1$, which equation represents the result of correctly applying differentiation rules to both sides with respect to $x$?
   Answer: $\tan(y) + x \sec^2(y) \frac{dy}{dx} = 2y \frac{dy}{dx}$ When you encounter an equation where both sides contain functions of different variables, you need to use implicit differentiation—differentiating both sides with respect to one variable while treating the other as a function of that variable. Starting with $x \tan(y) = y^2 + 1$, let's differentiate both sides with respect to $x$. On the left side, you have a product $x \tan(y)$, so apply the product rule: $\frac{d}{dx}[x \tan(y)] = 1 \cdot \tan(y) + x \cdot \frac{d}{dx}[\tan(y)]$. Since $y$ depends on $x$, you need the chain rule for $\frac{d}{dx}[\tan(y)] = \sec^2(y) \frac{dy}{dx}$. This gives you: $\tan(y) + x \sec^2(y) \frac{dy}{dx}$. On the right side, differentiate $y^2 + 1$: $\frac{d}{dx}[y^2 + 1] = 2y \frac{dy}{dx} + 0 = 2y \frac{dy}{dx}$. The complete differentiated equation is: $\tan(y) + x \sec^2(y) \frac{dy}{dx} = 2y \frac{dy}{dx}$, which is option B. Option A incorrectly omits the $\tan(y)$ term from the product rule. Option C forgets to include $\frac{dy}{dx}$ terms entirely—a fundamental error in implicit differentiation. Option D incorrectly differentiates the right side as $2y + \frac{dy}{dx}$ instead of $2y \frac{dy}{dx}$. Study tip: In implicit differentiation problems, always remember that every term containing the dependent variable needs a $\frac{dy}{dx}$ factor when you differentiate. Don't forget the product rule when variables are multiplied together!
3. The first derivative of the curve defined by $x^2 + y^2 = 25$ is $\frac{dy}{dx} = -\frac{x}{y}$. To find the second derivative $\frac{d^2y}{dx^2}$, the expression for $\frac{dy}{dx}$ is differentiated with respect to $x$. Which of the following statements correctly describes this procedure?
   Answer: Apply the quotient rule to $-\frac{x}{y}$, treating $y$ as a function of $x$ and using the chain rule to find its derivative, $\frac{dy}{dx}$. When finding the second derivative implicitly, we must differentiate the first derivative, $\frac{dy}{dx} = -\frac{x}{y}$, with respect to $x$. Since $y$ is implicitly a function of $x$, we cannot treat it as a constant (ruling out A). We must use the quotient rule, and when differentiating the $y$ in the denominator, we must use the chain rule, resulting in a $\frac{dy}{dx}$ term. This is correctly described in B. Option C is incorrect; differentiating $2x+2y\frac{dy}{dx}=0$ requires the product rule on the second term. Option D is a valid strategy for the top half of the circle, but it is not the required procedure for implicit differentiation and does not work for the entire curve without considering $y = -\sqrt{25-x^2}$ separately.
4. To find $\frac{dy}{dx}$ for the relation $\ln(x^3 y^2) = x+y$, which is the most efficient initial step?
   Answer: Apply properties of logarithms to get $3\ln(x) + 2\ln(y) = x+y$ and then differentiate implicitly. Using the properties of logarithms ($\ln(ab) = \ln(a) + \ln(b)$ and $\ln(a^n) = n\ln(a)$) to expand the left side first greatly simplifies the expression. Differentiating $3\ln(x) + 2\ln(y) = x+y$ is much more straightforward than differentiating the original expression, which would require a complicated chain rule and product rule combination. This makes it the most efficient first step.
5. To find the derivative of $f(x) = \sqrt{\frac{x+1}{x-1}}$, a student considers several different approaches. Which of the following statements is true?
   Answer: All three procedures—chain/quotient rule, direct quotient rule on square roots, and logarithmic differentiation—are valid. All three methods are valid procedures for finding the derivative. 1) Using the chain rule on $u^{1/2}$ and the quotient rule on the inner function $u = \frac{x+1}{x-1}$ is a standard approach. 2) Treating the function as $\frac{\sqrt{x+1}}{\sqrt{x-1}}$ and applying the quotient rule directly also works. 3) Logarithmic differentiation is also very effective: $\ln(f(x)) = \frac{1}{2}[\ln(x+1) - \ln(x-1)]$, and implicit differentiation yields the derivative. All paths lead to the same correct answer.
6. Let $f(x)$ be a differentiable function. To find the derivative of $y = [f(x)]^2$, one student uses the power rule combined with the chain rule. A second student uses the product rule on $y = f(x) \cdot f(x)$. Which statement accurately compares these procedures?
   Answer: Both procedures are valid and will yield the same result, $2f(x)f'(x)$. Both procedures are correct and demonstrate the consistency of differentiation rules. Using the chain rule (specifically, the generalized power rule): $\frac{d}{dx}[f(x)]^2 = 2[f(x)]^1 \cdot f'(x) = 2f(x)f'(x)$. Using the product rule on $f(x) \cdot f(x)$: $\frac{d}{dx}[f(x)f(x)] = f'(x)f(x) + f(x)f'(x) = 2f(x)f'(x)$. The results are identical.
7. Given the equation $x \ln(y) + y^3 = 2x$, which equation is the result of correctly differentiating with respect to $x$ to find $\frac{dy}{dx}$?
   Answer: $\ln(y) + \frac{x}{y}\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 2$ To find $\frac{dy}{dx}$ implicitly, we differentiate each term with respect to $x$. The term $x \ln(y)$ requires the product rule: the derivative is $(1) \cdot \ln(y) + x \cdot \frac{1}{y} \frac{dy}{dx}$. The term $y^3$ requires the chain rule: its derivative is $3y^2 \frac{dy}{dx}$. The derivative of $2x$ is $2$. Combining these results yields the equation in option B. Option A omits the $\frac{dy}{dx}$ factors that arise from the chain rule. Option C applies the product rule incorrectly to the first term. Option D also applies the product rule incorrectly, differentiating $x$ and $\ln(y)$ separately.
8. After finding $\frac{dy}{dx}$ for an implicitly defined curve such as $x^3 + y^3 = 6xy$, the resulting expression for the derivative typically involves both $x$ and $y$. What is the geometric reason for this?
   Answer: The slope of the tangent line at a point on the curve can depend on both the x- and y-coordinates of that point. When you encounter implicit differentiation problems, you're dealing with curves that can't easily be solved for y in terms of x alone. The key insight is understanding what the derivative $\frac{dy}{dx}$ geometrically represents: the slope of the tangent line at any point on the curve. The slope of a tangent line at a specific point depends on where you are on the curve - that is, it depends on both the x-coordinate and y-coordinate of that point. Consider the folium of Descartes $x^3 + y^3 = 6xy$: different points $(x,y)$ on this curve will generally have different slopes, and those slopes are determined by both coordinates of the point in question. Option C correctly captures this geometric reality - the slope at any point naturally depends on both coordinates of that point. Option A misses the point by focusing on solvability rather than geometry. Even when we can solve for y explicitly, implicit differentiation would still yield expressions involving both variables if we chose to use that method. Option B incorrectly suggests the chain rule automatically introduces extra y variables. The chain rule introduces $\frac{dy}{dx}$ terms, not additional y variables in the final answer. Option D confuses the nature of the derivative itself. The derivative $\frac{dy}{dx}$ is still fundamentally about how y changes with respect to x, but its value at any point happens to depend on both coordinates of that point. Remember: when you see both x and y in your implicit derivative, think "the slope depends on where I am on the curve."
9. A curve is defined by the equation $x^2y + y^3 = 10$. To find $\frac{dy}{dx}$, which of the following describes the necessary application of differentiation rules?
   Answer: Differentiate $x^2y$ using the product rule and $y^3$ using the chain rule, as part of implicit differentiation. The equation defines $y$ implicitly as a function of $x$. Therefore, implicit differentiation must be used. The term $x^2y$ is a product of two functions of $x$ (since $y$ is a function of $x$), so the product rule is required. The term $y^3$ is a composition of the cubing function and $y(x)$, so the chain rule is required, yielding $3y^2 \frac{dy}{dx}$.
10. For which of the following functions is it necessary to use logarithmic differentiation or a related method (like rewriting the base as $e^{\ln(\dots)}$) to find its derivative?
    Answer: $k(x) = (x+1)^{\sin x}$ Logarithmic differentiation is required for functions of the form $y=u(x)^{v(x)}$, where both the base $u(x)$ and the exponent $v(x)$ are non-constant functions of $x$. Option D, $k(x) = (x+1)^{\sin x}$, fits this form. Option A, $f(x) = (\pi)^x$, is an exponential function with a constant base, whose derivative is $(\ln \pi)(\pi)^x$. Option B, $g(x) = x^{\pi}$, is a power function with a constant exponent, whose derivative is $\pi x^{\pi-1}$. Option C, $h(x) = (\ln x)^2$, can be differentiated using the power rule and the chain rule.
11. A student must find the derivatives of three functions: I. $y = x^{\pi}$, II. $y = \pi^x$, III. $y = x^x$. Which choice correctly pairs the primary differentiation procedure for each function?
    Answer: I: Power Rule, II: Exponential Rule, III: Logarithmic Differentiation When you encounter functions with variables as bases and exponents, identifying the correct differentiation technique is crucial. The key is recognizing whether the base, exponent, or both contain variables. For function I, $y = x^{\pi}$, you have a variable base with a constant exponent. This is a perfect candidate for the Power Rule: $\frac{d}{dx}[x^n] = nx^{n-1}$, giving you $y' = \pi x^{\pi-1}$. Function II, $y = \pi^x$, has a constant base with a variable exponent. This is an exponential function requiring the Exponential Rule: $\frac{d}{dx}[a^x] = a^x \ln(a)$, yielding $y' = \pi^x \ln(\pi)$. Function III, $y = x^x$, has both variable base and variable exponent. Neither the Power Rule nor Exponential Rule applies directly. You need logarithmic differentiation: take the natural log of both sides, then differentiate implicitly to get $y' = x^x(\ln(x) + 1)$. This confirms answer C is correct. Answer A incorrectly suggests using logarithmic differentiation for $\pi^x$, which is unnecessary since it's a standard exponential function. Answer B misclassifies $\pi^x$ as needing the Power Rule and $x^x$ as needing the Power Rule, both wrong. Answer D incorrectly pairs the Chain Rule with $x^{\pi}$ and suggests implicit differentiation for $x^x$, though logarithmic differentiation is more direct. Study tip: Memorize this pattern: constant exponent → Power Rule, constant base → Exponential Rule, both variable → logarithmic differentiation.
12. The equation $y = \sec(x^2+y)$ defines a relation between $x$ and $y$. Which of the following describes a required method to begin finding $\frac{dy}{dx}$?
    Answer: Differentiate both sides with respect to $x$, which requires using implicit differentiation and applying the chain rule to the secant function. When you encounter an equation like $y = \sec(x^2+y)$ where the dependent variable $y$ appears on both sides, you're dealing with an implicit function that cannot be easily solved for $y$ explicitly. This signals that implicit differentiation is your primary tool. To find $\frac{dy}{dx}$, you must differentiate both sides of the equation with respect to $x$. The left side gives you $\frac{dy}{dx}$. For the right side, you need the chain rule: the derivative of $\sec(u)$ is $\sec(u)\tan(u) \cdot \frac{du}{dx}$, where $u = x^2+y$. Since $\frac{du}{dx} = 2x + \frac{dy}{dx}$, you get an equation containing $\frac{dy}{dx}$ terms that you can solve algebraically. This makes C correct. A is wrong because rewriting secant as $\frac{1}{\cos(u)}$ doesn't eliminate the implicit nature of the problem—you still have $y$ on both sides and would need implicit differentiation anyway. B is incorrect because taking $\text{arcsec}$ of both sides gives you $\text{arcsec}(y) = x^2+y$, which still has $y$ on both sides. You haven't simplified the problem at all. D is impossible because this equation cannot be solved algebraically for $y$. The transcendental nature of the secant function combined with $y$ appearing in its argument prevents explicit solution. Study tip: When you see the dependent variable appearing on both sides of an equation (especially inside transcendental functions), immediately think implicit differentiation—don't waste time trying to solve for the variable explicitly.
13. Let $f(x) = x^5 + 2x + 1$. If $g(x)$ is the inverse of $f(x)$, what procedure is required to find $g'(4)$?
    Answer: Find a value $c$ such that $f(c) = 4$, then calculate $f'(c)$, and finally compute the reciprocal $1/f'(c)$. The theorem for the derivative of an inverse function states that $(f^{-1})'(a) = \frac{1}{f'(f^{-1}(a))}$. Here, $g = f^{-1}$ and $a=4$. The first step is to find $c = f^{-1}(4)$, which means we must solve $f(c) = c^5 + 2c + 1 = 4$. By inspection, $c=1$. The next step is to find $f'(x) = 5x^4 + 2$. Finally, we evaluate $f'(c) = f'(1) = 5(1)^4 + 2 = 7$, and take the reciprocal. So, $g'(4) = 1/7$. Procedure B correctly outlines these steps.
14. Let $f$ be a differentiable, one-to-one function, and let $g(x) = f^{-1}(x)$. The tangent line to the graph of $y=f(x)$ at $x=2$ is $y = 5x - 7$. Which procedure leads to the value of $g'(3)$?
    Answer: Identify $f(2)=3$ and $f'(2)=5$ from the tangent line, then use these values in the formula $g'(3) = 1/f'(g(3))$. When you see inverse functions and derivatives together, you're dealing with the inverse function derivative theorem: if $g = f^{-1}$, then $g'(x) = \frac{1}{f'(g(x))}$. The key is identifying the right values to plug into this formula. From the tangent line $y = 5x - 7$ at $x = 2$, you can extract two crucial pieces of information: the slope is $f'(2) = 5$, and the point of tangency gives you $f(2) = 5(2) - 7 = 3$. Since $f(2) = 3$, this means $g(3) = 2$ (because $g$ undoes what $f$ does). Now you can apply the formula: $g'(3) = \frac{1}{f'(g(3))} = \frac{1}{f'(2)} = \frac{1}{5}$. This is exactly what option B describes. Option A makes the common error of thinking $g'(x)$ always equals $\frac{1}{f'(x)}$ at corresponding points, but this ignores that the domains are switched for inverse functions. Option C incorrectly treats the tangent line as if it were the actual function $f(x)$—the tangent line only approximates $f$ near $x = 2$, so its inverse isn't $g$. Option D uses $f'(3)$ instead of $f'(2)$, but we don't know $f'(3)$ and we need $f'$ evaluated at $g(3) = 2$, not at 3. Remember: for inverse function derivatives, you need both the derivative of the original function AND the correct input-output relationship between $f$ and $g$. Always identify these paired values from the given information first.
15. Let $f(x) = x^3 + x - 2$. To find $(f^{-1})'(0)$, what is the essential first computational step according to the inverse function theorem?
    Answer: Solve the equation $x^3 + x - 2 = 0$ to find the value of $f^{-1}(0)$. When you encounter questions about finding the derivative of an inverse function, the Inverse Function Theorem is your key tool. This theorem states that $(f^{-1})'(a) = \frac{1}{f'(f^{-1}(a))}$, meaning you need to know where the inverse function maps the input value before you can apply the formula. To find $(f^{-1})'(0)$, you must first determine what value of $x$ satisfies $f(x) = 0$. This is because $f^{-1}(0)$ represents the input that produces an output of 0 when plugged into $f(x)$. Solving $x^3 + x - 2 = 0$ gives you this crucial value. Once you know $f^{-1}(0) = a$, you can then calculate $f'(a)$ and apply the inverse function theorem formula. Option A is incorrect because finding critical points of $f(x)$ doesn't help you locate where $f(x) = 0$. Option B mistakenly evaluates the derivative at $x = 0$ rather than at $x = f^{-1}(0)$ – this confuses the input and output relationships. Option C attempts the most difficult approach by finding an explicit inverse formula, which is often impossible or unnecessarily complex for polynomial functions of degree 3 or higher. Remember this pattern: when finding $(f^{-1})'(a)$, your first step is always solving $f(x) = a$ to find the corresponding $x$-value. Only then can you evaluate $f'(x)$ at the correct point and apply the inverse function theorem.
16. A particle's position is given by $s(t) = \arcsin(\sqrt{t})$. To find the velocity $v(t) = s'(t)$, which procedure is correct?
    Answer: Apply the chain rule, using the derivative of $\arcsin(u)$ for the outer function and the derivative of $\sqrt{t}$ for the inner function. The function $s(t)$ is a composition of functions, $f(g(t))$, where the outer function is $f(u) = \arcsin(u)$ and the inner function is $g(t) = \sqrt{t}$. The correct procedure is to use the chain rule: $s'(t) = f'(g(t)) \cdot g'(t)$. The derivative of $\arcsin(u)$ is $\frac{1}{\sqrt{1-u^2}}$, and the derivative of $\sqrt{t}$ is $\frac{1}{2\sqrt{t}}$.
17. Consider the implicitly defined curve $y^2 - 2\ln(xy) = 1$. Which of the following describes the first step in the correct procedure to find the slope of the tangent line at a point $(x,y)$?
    Answer: Differentiate each term of the equation with respect to $x$, applying the chain and product rules where necessary. When you encounter an implicitly defined equation like $y^2 - 2\ln(xy) = 1$, you're dealing with a curve where y isn't explicitly solved for in terms of x. To find the slope (which is $\frac{dy}{dx}$), you need implicit differentiation. The correct first step is to differentiate both sides of the equation with respect to x, which is exactly what option C describes. As you differentiate each term, you'll need the chain rule for terms involving y (since y depends on x), and the product rule for terms like $\ln(xy)$. For example, differentiating $y^2$ gives $2y\frac{dy}{dx}$, and differentiating $\ln(xy)$ requires both the chain rule and product rule, yielding $\frac{1}{xy}(y + x\frac{dy}{dx})$. Option A is impossible here because this equation cannot be solved explicitly for y in terms of x due to the logarithmic term. Option B suggests differentiating with respect to y to find $\frac{dx}{dy}$, but the question asks for the slope $\frac{dy}{dx}$, not its reciprocal. While you could find $\frac{dx}{dy}$ and then take its reciprocal, this creates unnecessary work. Option D puts the cart before the horse—you need the general derivative formula first before you can evaluate it at any specific point. Study tip: Whenever you see an equation that mixes x and y terms (especially with transcendental functions like ln), think "implicit differentiation." Always differentiate with respect to x first, then solve for $\frac{dy}{dx}$.
18. To find the derivative of the function $f(x) = e^{\sin(x^2)}$, which differentiation rule must be applied as the primary, outermost step?
    Answer: The chain rule, because the function is a composition of an exponential, a sine, and a power function. When you encounter a function like $f(x) = e^{\sin(x^2)}$, you need to identify its structure to choose the right differentiation approach. This function is a composition - an exponential function containing a sine function, which itself contains a quadratic function. The chain rule is your primary tool here because you're differentiating a composite function. Think of it as peeling an onion: the outermost layer is $e^u$ (where $u = \sin(x^2)$), the middle layer is $\sin(v)$ (where $v = x^2$), and the innermost layer is $x^2$. To find $f'(x)$, you'll need: $f'(x) = e^{\sin(x^2)} \cdot \cos(x^2) \cdot 2x$ This uses the chain rule three times, working from outside to inside. Looking at the wrong answers: Choice A suggests the product rule, but there's no multiplication of separate functions here - just one composite function. Choice B mentions applying the exponential rule directly to the argument, but you can't ignore that the argument itself is a complex function requiring its own differentiation. Choice C focuses on the power rule for $x^2$, but while the power rule will eventually be used (as the innermost step), it's not the primary approach needed. Study tip: When you see nested functions like $f(g(h(x)))$, always start with the chain rule as your primary strategy. Look for the outermost function first, then work your way inward. The chain rule often combines with other rules, but it governs the overall approach.
19. To find the derivative of $y = \cos^3(\ln(x^2))$, what is the most appropriate sequence of differentiation rules to apply for the main functional structure?
    Answer: Power rule for the exponent 3, then chain rule for cosine, then chain rule for the natural log. The function is a composition of several functions. The outermost operation is cubing, i.e., $u^3$ where $u = \cos(\ln(x^2))$. So, the first rule is the power rule. The next inner function is $\cos(v)$ where $v = \ln(x^2)$, which requires the chain rule. The innermost function is $\ln(w)$ where $w=x^2$, which requires another application of the chain rule. Thus, the sequence is Power Rule, then Chain Rule, then Chain Rule again.
20. Let $f(x) = x^3 + 2x - 1$. If $g(x) = f^{-1}(x)$, what is the value of $g'(2)$?
    Answer: $\frac{1}{5}$ The formula for the derivative of an inverse function is $(f^{-1})'(a) = \frac{1}{f'(f^{-1}(a))}$. Here, $a=2$. First, we must find $b = f^{-1}(2)$, which is the value of $x$ such that $f(x)=2$. We solve $x^3 + 2x - 1 = 2$, which simplifies to $x^3 + 2x - 3 = 0$. By inspection, $x=1$ is the solution, so $f^{-1}(2)=1$. Next, we find the derivative of $f(x)$: $f'(x) = 3x^2 + 2$. Now we evaluate $f'(f^{-1}(2)) = f'(1) = 3(1)^2 + 2 = 5$. Finally, $g'(2) = \frac{1}{5}$. Distractor D is $1/f'(2)$, a common error. Distractor C is $f'(1)$ without taking the reciprocal. Distractor A is $1/f(2)$.

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