Higher-Order Derivatives - Calculus 1

If $y = \arctan(x)$ , find an expression for $\frac{d^2y}{dx^2}$ .
Answer: $-\frac{2x}{(1+x^2)^2}$ When finding the second derivative of inverse trigonometric functions, you'll need to apply the chain rule and quotient rule systematically. This type of problem tests your ability to differentiate composite functions multiple times. Starting with $y = \arctan(x)$ , you first need the first derivative. The derivative of $\arctan(x)$ is $\frac{dy}{dx} = \frac{1}{1+x^2}$ . Now you need to differentiate this expression to find $\frac{d^2y}{dx^2}$ . Since $\frac{dy}{dx} = (1+x^2)^{-1}$ , use the chain rule: $\frac{d^2y}{dx^2} = -1 \cdot (1+x^2)^{-2} \cdot \frac{d}{dx}(1+x^2) = -(1+x^2)^{-2} \cdot 2x = -\frac{2x}{(1+x^2)^2}$ . This confirms answer C is correct. Let's examine why the other choices are wrong. Choice A, $-\frac{2x}{1+x^2}$ , represents what you'd get if you incorrectly applied the power rule without squaring the denominator—forgetting that differentiating $(1+x^2)^{-1}$ requires the chain rule. Choice B, $\frac{1}{(1+x^2)^2}$ , has the right denominator but wrong sign and numerator, suggesting confusion about the chain rule's negative contribution. Choice D, $-\frac{1}{(1+x^2)^2}$ , gets the sign right but misses the $2x$ factor from differentiating the inner function $1+x^2$ . Remember: when differentiating inverse trig functions twice, the second derivative will typically involve the derivative of the "inside" function (here, $2x$ ) appearing in the numerator due to the chain rule.
Given that $y' = x^2 + y^2$ , find $y''$ at the point $(1, 1)$ .
Answer: 6 We are given $y' = x^2 + y^2$ . To find $y''$ , we differentiate this equation with respect to $x$ : $y'' = 2x + 2y \cdot y'$ . We need to evaluate this at the point $(1, 1)$ . First, we find the value of $y'$ at $(1, 1)$ using the given equation: $y'(1,1) = 1^2 + 1^2 = 2$ . Now we can find $y''$ at $(1, 1)$ by substituting $x=1$ , $y=1$ , and $y'=2$ into the expression for $y''$ : $y''(1,1) = 2(1) + 2(1)(2) = 2 + 4 = 6$ .
Let $f(x) = \frac{x}{x+1}$ . Find $f^{(10)}(1)$ .
Answer: $-\frac{10!}{2^{11}}$ First, rewrite $f(x) = \frac{x+1-1}{x+1} = 1 - \frac{1}{x+1} = 1 - (x+1)^{-1}$ . The derivative of the constant 1 is zero, so we only need to find the 10th derivative of $-(x+1)^{-1}$ . Let $g(x) = -(x+1)^{-1}$ . $g'(x) = (x+1)^{-2}$ , $g''(x) = -2(x+1)^{-3}$ , $g'''(x) = 6(x+1)^{-4}$ . The pattern is $g^{(n)}(x) = (-1)^{n+1} n! (x+1)^{-(n+1)}$ . For $n=10$ , we have $f^{(10)}(x) = g^{(10)}(x) = (-1)^{11} 10! (x+1)^{-11} = -\frac{10!}{(x+1)^{11}}$ . Evaluating at $x=1$ gives $f^{(10)}(1) = -\frac{10!}{(1+1)^{11}} = -\frac{10!}{2^{11}}$ .
Let $f(x) = e^{x^2}$ . What is the value of $f''(2)$ ?
Answer: $18e^4$ First, find the first derivative using the chain rule: $f'(x) = e^{x^2} \cdot 2x = 2xe^{x^2}$ . Next, find the second derivative using the product rule and chain rule: $f''(x) = (2)(e^{x^2}) + (2x)(e^{x^2} \cdot 2x) = 2e^{x^2} + 4x^2e^{x^2} = (2 + 4x^2)e^{x^2}$ . Finally, evaluate at $x=2$ : $f''(2) = (2 + 4(2^2))e^{2^2} = (2 + 16)e^4 = 18e^4$ .
If $e^y + y = x$ , what is $\frac{d^2y}{dx^2}$ in terms of $y$ ?
Answer: $-\frac{e^y}{(e^y+1)^3}$ This problem tests implicit differentiation and finding second derivatives. When you have an equation like $e^y + y = x$ where y is implicitly defined as a function of x, you need to differentiate twice to find the second derivative. First, find $\frac{dy}{dx}$ by differentiating both sides with respect to x: $\frac{d}{dx}(e^y + y) = \frac{d}{dx}(x)$ $e^y \frac{dy}{dx} + \frac{dy}{dx} = 1$ $(e^y + 1)\frac{dy}{dx} = 1$ $\frac{dy}{dx} = \frac{1}{e^y + 1}$ Now differentiate this result to find the second derivative: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{1}{e^y + 1}\right)$ Using the chain rule on this quotient: $\frac{d^2y}{dx^2} = -\frac{1}{(e^y + 1)^2} \cdot \frac{d}{dx}(e^y + 1) = -\frac{1}{(e^y + 1)^2} \cdot e^y \frac{dy}{dx}$ Substituting $\frac{dy}{dx} = \frac{1}{e^y + 1}$ : $\frac{d^2y}{dx^2} = -\frac{e^y}{(e^y + 1)^2} \cdot \frac{1}{e^y + 1} = -\frac{e^y}{(e^y + 1)^3}$ This matches answer B. Answer A has the wrong sign and wrong power in the denominator. Answer C is missing the $e^y$ factor in the numerator—this would be the result if you forgot to apply the chain rule when differentiating $e^y + 1$ . Answer D has the correct expression but wrong sign. Remember: when finding second derivatives of implicit functions, carefully track your chain rule applications and substitute your first derivative back into the expression.
Let $p(x)$ be a polynomial of degree 4. What is the value of $p^{(5)}(x)$ , the fifth derivative of $p(x)$ ?
Answer: 0 When you encounter questions about derivatives of polynomials, think systematically about how differentiation affects polynomial degree. Each time you differentiate a polynomial, the degree decreases by exactly 1. Let's work through this step-by-step. A degree 4 polynomial has the general form $p(x) = ax^4 + bx^3 + cx^2 + dx + e$ where $a \neq 0$ . Now let's find each successive derivative: - $p'(x) = 4ax^3 + 3bx^2 + 2cx + d$ (degree 3) - $p''(x) = 12ax^2 + 6bx + 2c$ (degree 2) - $p'''(x) = 24ax + 6b$ (degree 1) - $p^{(4)}(x) = 24a$ (degree 0, a constant) - $p^{(5)}(x) = 0$ (no terms left to differentiate) Once you've differentiated enough times to eliminate all variable terms, any further derivatives are zero. Choice A is wrong because the fifth derivative has no terms with $x$ —we eliminated the last variable term when going from the fourth to fifth derivative. Choice B is incorrect because while the fourth derivative is indeed a non-zero constant, the fifth derivative of that constant is zero. Choice D misses the fundamental point that regardless of the specific coefficients, the pattern of degree reduction is always the same. Study tip: Remember the rule: if you differentiate a polynomial $n+1$ times where $n$ is its degree, you'll always get zero. This works for any polynomial, regardless of coefficients.
If $y^3 - xy = 6$ , find the value of $\frac{d^2y}{dx^2}$ at the point $(1, 2)$ .
Answer: $-\frac{4}{1331}$ First, differentiate implicitly: $3y^2 y' - (1\cdot y + x y') = 0$ , which gives $y'(3y^2 - x) = y$ , so $y' = \frac{y}{3y^2 - x}$ . At $(1, 2)$ , $y' = \frac{2}{3(2^2)-1} = \frac{2}{11}$ . Differentiate the equation $3y^2 y' - y - xy' = 0$ again: $(6y y')y' + 3y^2 y'' - y' - (y' + x y'') = 0$ . Rearranging gives $y''(3y^2-x) = 2y' - 6y(y')^2$ . At $(1, 2)$ , we have $y''(3(2^2)-1) = 2(\frac{2}{11}) - 6(2)(\frac{2}{11})^2$ , so $11y'' = \frac{4}{11} - \frac{48}{121} = \frac{44-48}{121} = -\frac{4}{121}$ . Thus, $y'' = -\frac{4}{1331}$ .
The position of a particle moving along a line is given by $s(t) = t^2 e^{-t}$ for $t \ge 0$ . The jerk of the particle is defined as the third derivative of position, $s'''(t)$ . What is the smallest time $t>0$ at which the particle's jerk is zero?
Answer: $3 - \sqrt{3}$ When you encounter problems involving jerk (the third derivative of position), you need to systematically take derivatives and then solve for when that third derivative equals zero. Starting with $s(t) = t^2 e^{-t}$ , you'll need to find $s'''(t)$ . For the first derivative, use the product rule: $s'(t) = 2t e^{-t} + t^2(-e^{-t}) = e^{-t}(2t - t^2)$ For the second derivative, apply the product rule again: $s''(t) = -e^{-t}(2t - t^2) + e^{-t}(2 - 2t) = e^{-t}(t^2 - 4t + 2)$ For the third derivative: $s'''(t) = -e^{-t}(t^2 - 4t + 2) + e^{-t}(2t - 4) = e^{-t}(-t^2 + 6t - 6)$ Setting $s'''(t) = 0$ : Since $e^{-t} \neq 0$ , you need $-t^2 + 6t - 6 = 0$ , or $t^2 - 6t + 6 = 0$ . Using the quadratic formula: $t = \frac{6 \pm \sqrt{36 - 24}}{2} = \frac{6 \pm 2\sqrt{3}}{2} = 3 \pm \sqrt{3}$ Since you want the smallest positive time, choose $t = 3 - \sqrt{3}$ (answer D). Answer A ( $1 + \sqrt{3} \approx 2.73$ ) and answer C ( $2$ ) are smaller values that don't satisfy our equation. Answer B ( $2 - \sqrt{2} \approx 0.59$ ) is the smallest numerically but also doesn't solve our quadratic equation. Remember: with higher-order derivatives of products involving exponentials, always factor out the exponential term before setting equal to zero—it simplifies the algebra significantly.
The $n$ -th derivative of $f(x) = x e^x$ is given by $f^{(n)}(x) = (x+n)e^x$ . Using this fact, what is $f^{(100)}(0)$ ?
Answer: 100 When you encounter a problem that gives you a formula for the n-th derivative, you're being tested on your ability to apply that formula directly and evaluate derivatives at specific points. Given that $f^{(n)}(x) = (x+n)e^x$ , you can find the 100th derivative by substituting $n = 100$ : $f^{(100)}(x) = (x+100)e^x$ . To evaluate this at $x = 0$ , substitute: $f^{(100)}(0) = (0+100)e^0 = 100 \cdot 1 = 100$ . Let's examine why the other answers are incorrect. Choice (A) gives 0, which would only be true if the coefficient of $e^x$ were zero - but $(0+100) = 100 \neq 0$ . Choice (B) suggests 100!, which might seem reasonable since factorials often appear in derivative formulas, but this specific formula $(x+n)e^x$ doesn't involve factorials when evaluated at zero. Choice (C) gives 1, which you might get if you forgot about the $(x+n)$ term and only considered $e^0 = 1$ , but you can't ignore the coefficient. The key insight is recognizing that when $x = 0$ , the exponential term $e^0$ always equals 1, so the entire expression simplifies to just the coefficient $(0+n) = n$ . This makes the calculation straightforward: the 100th derivative of $xe^x$ evaluated at zero is simply 100. Study tip: When evaluating derivatives at $x = 0$ , remember that $e^0 = 1$ , which often simplifies complex exponential expressions to just their coefficients.
What is the rate of change of the slope of the tangent line to the curve $y = x \cos(x)$ at $x=\frac{\pi}{2}$ ?
Answer: -2 This question tests your understanding of second derivatives and their geometric meaning. When you're asked about the "rate of change of the slope," you need to recognize that slope itself is the first derivative, so the rate of change of slope is the second derivative. To find the second derivative of $y = x \cos(x)$ , start with the first derivative using the product rule: $y' = \frac{d}{dx}[x \cos(x)] = 1 \cdot \cos(x) + x \cdot (-\sin(x)) = \cos(x) - x\sin(x)$ Now differentiate again to get the second derivative: $y'' = \frac{d}{dx}[\cos(x) - x\sin(x)] = -\sin(x) - [1 \cdot \sin(x) + x \cdot \cos(x)] = -\sin(x) - \sin(x) - x\cos(x) = -2\sin(x) - x\cos(x)$ Evaluating at $x = \frac{\pi}{2}$ : $y''\left(\frac{\pi}{2}\right) = -2\sin\left(\frac{\pi}{2}\right) - \frac{\pi}{2}\cos\left(\frac{\pi}{2}\right) = -2(1) - \frac{\pi}{2}(0) = -2$ So choice B) -2 is correct. Choice A) $\pi$ might result from confusing the evaluation point with the answer. Choice C) 1 could come from evaluating only $\sin(\frac{\pi}{2})$ without the coefficient. Choice D) 0 might result from incorrectly thinking that $\cos(\frac{\pi}{2}) = 0$ dominates the calculation. Remember: "rate of change of slope" always means second derivative. Master the product rule for both first and second derivatives, as functions involving products appear frequently in calculus.
A curve is defined by the parametric equations $x(t) = t^3 - 3t$ and $y(t) = t^2 - 2$ . Find the value of $\frac{d^2y}{dx^2}$ at the point where $t=2$ .
Answer: $-\frac{10}{243}$ First, we find the derivatives with respect to $t$ : $\frac{dx}{dt} = 3t^2 - 3$ and $\frac{dy}{dt} = 2t$ . Then, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t}{3t^2-3}$ . The second derivative is $\frac{d^2y}{dx^2} = \frac{d/dt(dy/dx)}{dx/dt}$ . The numerator is $\frac{d}{dt}(\frac{2t}{3t^2-3}) = \frac{2(3t^2-3) - 2t(6t)}{(3t^2-3)^2} = \frac{-6t^2-6}{(3t^2-3)^2}$ . So, $\frac{d^2y}{dx^2} = \frac{-6t^2-6}{(3t^2-3)^3}$ . At $t=2$ , $\frac{d^2y}{dx^2} = \frac{-6(2^2)-6}{(3(2^2)-3)^3} = \frac{-30}{9^3} = \frac{-30}{729} = -\frac{10}{243}$ .
Let $f(x) = \ln(1-x)$ . Find the value of $f^{(5)}(0)$ , the fifth derivative of $f$ evaluated at $x=0$ .
Answer: -24 Finding higher-order derivatives of logarithmic functions requires recognizing patterns and applying systematic differentiation. When you encounter $f(x) = \ln(1-x)$ , the key is to find the derivatives step by step and look for a pattern. Let's compute the derivatives systematically: - $f(x) = \ln(1-x)$ - $f'(x) = \frac{-1}{1-x} = -(1-x)^{-1}$ - $f''(x) = -(1)(1-x)^{-2} \cdot (-1) = -(1-x)^{-2}$ - $f'''(x) = -(-2)(1-x)^{-3} \cdot (-1) = -2(1-x)^{-3}$ - $f^{(4)}(x) = -2(-3)(1-x)^{-4} \cdot (-1) = -6(1-x)^{-4}$ - $f^{(5)}(x) = -6(-4)(1-x)^{-5} \cdot (-1) = -24(1-x)^{-5}$ Evaluating at $x = 0$ : $f^{(5)}(0) = -24(1-0)^{-5} = -24$ , confirming answer (D). The wrong answers represent common calculation errors: (A) 120 likely comes from forgetting the negative sign and miscalculating the factorial pattern. (B) 24 results from dropping the negative sign entirely. (C) -120 suggests correctly identifying the negative sign but incorrectly computing $4! = 24$ as $5! = 120$ . Study tip: For derivatives of $\ln(1-x)$ , the $n$ -th derivative follows the pattern $f^{(n)}(x) = -(n-1)!(1-x)^{-n}$ . Always track your negative signs carefully and remember that the factorial is $(n-1)!$ , not $n!$ .
Let $f(x) = \sin(2x)$ . Find $f^{(15)}(x)$ , the 15th derivative of $f(x)$ .
Answer: $-2^{15}\cos(2x)$ The derivatives of $\sin(2x)$ follow a cycle of four: $f'(x) = 2\cos(2x)$ , $f''(x) = -4\sin(2x)$ , $f'''(x) = -8\cos(2x)$ , $f^{(4)}(x) = 16\sin(2x)$ . The general pattern involves a factor of $2^n$ and a repeating trigonometric part. To find the 15th derivative, we find the remainder of $15$ divided by $4$ , which is $3$ (since $15 = 4 \cdot 3 + 3$ ). Therefore, the 15th derivative will follow the pattern of the 3rd derivative. The 3rd derivative is $-2^3\cos(2x)$ , so the 15th derivative is $f^{(15)}(x) = -2^{15}\cos(2x)$ .
Let $h(x) = e^{-x} \sin(2x)$ . What is the value of $h''(0)$ ?
Answer: $-4$ When you need to find the second derivative of a product involving exponential and trigonometric functions, the product rule becomes your essential tool. For $h(x) = e^{-x} \sin(2x)$ , you'll need to apply it twice. First, find $h'(x)$ using the product rule: $h'(x) = (e^{-x})' \sin(2x) + e^{-x} (\sin(2x))'$ . The derivative of $e^{-x}$ is $-e^{-x}$ , and the derivative of $\sin(2x)$ is $2\cos(2x)$ . So $h'(x) = -e^{-x}\sin(2x) + e^{-x} \cdot 2\cos(2x) = e^{-x}(-\sin(2x) + 2\cos(2x))$ . Now apply the product rule again to find $h''(x)$ : $h''(x) = (e^{-x})'(-\sin(2x) + 2\cos(2x)) + e^{-x}(-\sin(2x) + 2\cos(2x))'$ This gives us: $h''(x) = -e^{-x}(-\sin(2x) + 2\cos(2x)) + e^{-x}(-2\cos(2x) - 4\sin(2x))$ Simplifying: $h''(x) = e^{-x}(\sin(2x) - 2\cos(2x) - 2\cos(2x) - 4\sin(2x)) = e^{-x}(-3\sin(2x) - 4\cos(2x))$ At $x = 0$ : $h''(0) = e^0(-3\sin(0) - 4\cos(0)) = 1(-3 \cdot 0 - 4 \cdot 1) = -4$ . This confirms answer C. Answer A (0) would result from forgetting the cosine terms. Answer B (-2) likely comes from errors in applying the product rule or chain rule. Answer D (2) represents sign errors in the differentiation process. Remember: when differentiating products repeatedly, organize your work carefully and double-check each application of the product rule to avoid computational errors.
Let $g(x) = x \ln(x)$ . What is the value of $g^{(4)}(1)$ , the fourth derivative of $g(x)$ evaluated at $x=1$ ?
Answer: $2$ When you encounter a problem asking for a higher-order derivative, you need to systematically compute each derivative step by step until you reach the desired order. Starting with $g(x) = x \ln(x)$ , let's find each derivative using the product rule where needed: First derivative: $g'(x) = \frac{d}{dx}[x \ln(x)] = 1 \cdot \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1$ Second derivative: $g''(x) = \frac{d}{dx}[\ln(x) + 1] = \frac{1}{x} + 0 = \frac{1}{x}$ Third derivative: $g'''(x) = \frac{d}{dx}\left[\frac{1}{x}\right] = \frac{d}{dx}[x^{-1}] = -x^{-2} = -\frac{1}{x^2}$ Fourth derivative: $g^{(4)}(x) = \frac{d}{dx}\left[-\frac{1}{x^2}\right] = \frac{d}{dx}[-x^{-2}] = -(-2)x^{-3} = \frac{2}{x^3}$ Now evaluating at $x = 1$ : $g^{(4)}(1) = \frac{2}{1^3} = 2$ Answer choice A) $-1$ might come from confusing signs or stopping at the third derivative and evaluating incorrectly. Answer choice B) 1 could result from missing the coefficient 2 in the final derivative. Answer choice D) $-2$ represents getting the magnitude right but making a sign error when differentiating $-\frac{1}{x^2}$ . Study tip: For higher-order derivatives, work methodically one step at a time and double-check your signs. Pattern recognition helps too—derivatives of $x^n$ follow predictable factorial-like patterns.
For the curve defined by the equation $x^3 + y^3 = 9$ , what is the value of $\frac{d^2y}{dx^2}$ at the point $(1, 2)$ ?
Answer: $-\frac{9}{16}$ First, we find the first derivative using implicit differentiation: $3x^2 + 3y^2 \frac{dy}{dx} = 0$ , which simplifies to $\frac{dy}{dx} = -\frac{x^2}{y^2}$ . At the point $(1, 2)$ , we have $\frac{dy}{dx} = -\frac{1^2}{2^2} = -\frac{1}{4}$ . Next, we differentiate $\frac{dy}{dx}$ with respect to $x$ using the quotient rule to find the second derivative: $\frac{d^2y}{dx^2} = -\frac{(2x)(y^2) - (x^2)(2y \frac{dy}{dx})}{ (y^2)^2 } = -\frac{2xy^2 - 2x^2y \frac{dy}{dx}}{y^4}$ Now, we substitute the values from the point $(1, 2)$ and the value of $\frac{dy}{dx} = -\frac{1}{4}$ : $\frac{d^2y}{dx^2} = -\frac{2(1)(2^2) - 2(1^2)(2)(-\frac{1}{4})}{2^4} = -\frac{8 - 4(-\frac{1}{4})}{16} = -\frac{8 - (-1)}{16} = -\frac{8+1}{16} = -\frac{9}{16}$
Let $f(x) = \frac{1}{2x+3}$ . Which of the following is an expression for $f^{(n)}(x)$ , the $n$ -th derivative of $f(x)$ , for any integer $n \ge 1$ ?
Answer: $f^{(n)}(x) = \frac{(-1)^n n! 2^n}{(2x+3)^{n+1}}$ We write $f(x) = (2x+3)^{-1}$ and compute the first few derivatives to find a pattern: $f'(x) = -1(2x+3)^{-2} \cdot 2 = (-1)^1 \cdot 1! \cdot 2^1 \cdot (2x+3)^{-2}$ $f''(x) = (-1)( -2)(2x+3)^{-3} \cdot 2 \cdot 2 = (-1)^2 \cdot 2! \cdot 2^2 \cdot (2x+3)^{-3}$ $f'''(x) = (-1)^2 \cdot 2! \cdot (-3)(2x+3)^{-4} \cdot 2 \cdot 2^2 = (-1)^3 \cdot 3! \cdot 2^3 \cdot (2x+3)^{-4}$ Observing the pattern, the $n$ -th derivative has a factor of $(-1)^n$ , a factorial of $n!$ , a power of $2^n$ from repeated application of the chain rule, and the denominator term is $(2x+3)$ raised to the power of $-(n+1)$ . Thus, the general formula is $f^{(n)}(x) = (-1)^n n! 2^n (2x+3)^{-(n+1)} = \frac{(-1)^n n! 2^n}{(2x+3)^{n+1}}$ .
Let $f(x) = \cos(3x)$ . What is $f^{(2023)}(x)$ ?
Answer: $3^{2023}\sin(3x)$ When finding high-order derivatives of trigonometric functions, look for patterns that repeat cyclically. This saves you from computing thousands of derivatives individually. Let's find the pattern for $f(x) = \cos(3x)$ . Using the chain rule: - $f'(x) = -3\sin(3x)$ - $f''(x) = -3^2\cos(3x)$ - $f'''(x) = 3^3\sin(3x)$ - $f^{(4)}(x) = 3^4\cos(3x)$ Notice the pattern repeats every 4 derivatives! The coefficients follow powers of 3, while the trigonometric parts cycle: $\cos \to -\sin \to -\cos \to \sin \to \cos$ . To find $f^{(2023)}(x)$ , divide 2023 by 4: $2023 = 4 \times 505 + 3$ . Since the remainder is 3, $f^{(2023)}(x)$ has the same form as $f'''(x) = 3^3\sin(3x)$ , but with coefficient $3^{2023}$ . Therefore, $f^{(2023)}(x) = 3^{2023}\sin(3x)$ . Answer A gives the wrong trigonometric function (cosine instead of sine). Answer B has the correct sine function but includes an incorrect negative sign. Answer C combines both errors: wrong function (cosine) and wrong sign (negative). Only answer D correctly identifies both the sine function and positive coefficient. Study tip: For derivatives of $\sin(kx)$ and $\cos(kx)$ , the pattern always repeats every 4 derivatives. Find the remainder when dividing the derivative order by 4, then match it to your pattern. This technique works for any trigonometric function with a linear argument.
Given the curve defined by $x = \sin(t)$ and $y = \cos(2t)$ , find $\frac{d^2y}{dx^2}$ in terms of $x$ .
Answer: -4 When you encounter parametric equations and need to find the second derivative $\frac{d^2y}{dx^2}$ , you're working with the chain rule in a multi-step process. The key insight is that you'll need to find $\frac{dy}{dx}$ first, then differentiate that with respect to $x$ . Starting with $x = \sin(t)$ and $y = \cos(2t)$ , find the first derivatives: $\frac{dx}{dt} = \cos(t)$ and $\frac{dy}{dt} = -2\sin(2t)$ . Using the chain rule, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-2\sin(2t)}{\cos(t)}$ . Since $\sin(2t) = 2\sin(t)\cos(t)$ , this becomes $\frac{dy}{dx} = \frac{-4\sin(t)\cos(t)}{\cos(t)} = -4\sin(t)$ . Because $x = \sin(t)$ , we have $\frac{dy}{dx} = -4x$ . For the second derivative, differentiate $\frac{dy}{dx} = -4x$ with respect to $x$ : $\frac{d^2y}{dx^2} = -4$ . Answer choice A) $2x$ incorrectly suggests the second derivative depends on $x$ , likely from confusing first and second derivatives. Answer choice B) $-2$ gets the sign right but misses a factor of 2, probably from forgetting the coefficient in $\cos(2t)$ . Answer choice D) $-4x$ is actually $\frac{dy}{dx}$ , not $\frac{d^2y}{dx^2}$ – a common error of stopping one step too early. Remember: with parametric second derivatives, work systematically through each step and always check whether your final answer should depend on the parameter or be constant.
For what value of $k$ does $y=e^{kx}$ satisfy the equation $y'' + y' - 6y = 0$ ?
Answer: $k=2$ or $k=-3$ When you encounter a differential equation with an exponential function guess, you're dealing with a characteristic equation problem. The key insight is that exponential functions maintain their form through differentiation, making them perfect candidates for linear differential equations. Let's substitute $y = e^{kx}$ into the equation $y'' + y' - 6y = 0$ . First, find the derivatives: $y' = ke^{kx}$ and $y'' = k^2e^{kx}$ . Substituting these into the differential equation gives us: $k^2e^{kx} + ke^{kx} - 6e^{kx} = 0$ Factor out $e^{kx}$ (which is never zero): $e^{kx}(k^2 + k - 6) = 0$ This means $k^2 + k - 6 = 0$ . This is called the characteristic equation. Factoring: $(k + 3)(k - 2) = 0$ , so $k = -3$ or $k = 2$ . Looking at the wrong answers: Choice A gives $k = 3$ or $k = -2$ , which would satisfy $k^2 - k - 6 = 0$ (wrong signs in the original equation). Choice C gives $k = 6$ or $k = -1$ , satisfying $k^2 - 5k - 6 = 0$ (incorrect constant term). Choice D gives $k = 1$ or $k = -6$ , satisfying $k^2 + 5k - 6 = 0$ (wrong middle coefficient). Study tip: Always set up the characteristic equation by substituting $y = e^{rx}$ and its derivatives, then factor out the exponential term. The roots of the resulting polynomial give you the values that make the exponential function a solution.

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