Implicit Differentiation - Calculus 1

1. Let $y$ be a function of $x$ defined implicitly by $[f(y)]^2 = x^2y + 7$, where $f$ is a differentiable function. If $f(2)=3$ and $f'(2)=\frac{1}{2}$, what is the value of $\frac{dy}{dx}$ at the point $(1, 2)$?
   Answer: $2$ First, differentiate the equation $[f(y)]^2 = x^2y + 7$ implicitly with respect to $x$. Using the chain rule on the left side and the product rule on the right side, we get: $2f(y) \cdot f'(y) \cdot \frac{dy}{dx} = (2xy + x^2 \frac{dy}{dx}) + 0$. Now, substitute the given values: the point is $(x,y) = (1,2)$, and at this point $f(2)=3$ and $f'(2)=\frac{1}{2}$. The equation becomes: $2f(2) \cdot f'(2) \cdot \frac{dy}{dx} = 2(1)(2) + (1)^2 \frac{dy}{dx}$. Plugging in the function values: $2(3)(\frac{1}{2}) \frac{dy}{dx} = 4 + \frac{dy}{dx}$. This simplifies to $3\frac{dy}{dx} = 4 + \frac{dy}{dx}$. Solving for $\frac{dy}{dx}$ gives $2\frac{dy}{dx} = 4$, so $\frac{dy}{dx} = 2$.
2. What is the $y$-intercept of the normal line to the curve $x^2 - xy + y^2 = 3$ at the point $(-1, 1)$?
   Answer: $0$ When you encounter a problem asking for the y-intercept of a normal line to a curve, you're dealing with implicit differentiation and perpendicular lines. The normal line is perpendicular to the tangent line at the given point. First, find the slope of the tangent line by implicitly differentiating $x^2 - xy + y^2 = 3$. Taking the derivative with respect to $x$: $2x - y - x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$. Solving for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{2x - y}{x - 2y}$. At the point $(-1, 1)$, the slope of the tangent line is $\frac{dy}{dx} = \frac{2(-1) - 1}{(-1) - 2(1)} = \frac{-3}{-3} = 1$. Since the normal line is perpendicular to the tangent line, its slope is the negative reciprocal: $m_{normal} = -1$. Using point-slope form with the point $(-1, 1)$ and slope $-1$: $y - 1 = -1(x - (-1))$, which simplifies to $y - 1 = -(x + 1)$, or $y = -x$. The y-intercept occurs when $x = 0$, giving $y = 0$. Choice A) $-1$ would result from using the x-coordinate as the y-intercept. Choice B) 2 might come from incorrectly calculating the slope or using the wrong point-slope formula. Choice C) 1 would result from using the y-coordinate of the given point as the y-intercept. Remember: for normal line problems, always find the tangent slope first through implicit differentiation, then use the negative reciprocal for the normal slope. The y-intercept is found by setting $x = 0$ in your final equation.
3. Consider the curve defined by $x^3 - y^3 = 7$. Find the value of the second derivative $\frac{d^2y}{dx^2}$ at the point $(2, 1)$.
   Answer: $-28$ When you encounter implicit differentiation problems asking for the second derivative, you need to find the first derivative, then differentiate it again while carefully applying the chain rule. Starting with $x^3 - y^3 = 7$, differentiate both sides with respect to $x$: $3x^2 - 3y^2\frac{dy}{dx} = 0$ Solving for the first derivative: $\frac{dy}{dx} = \frac{x^2}{y^2}$ At point $(2,1)$: $\frac{dy}{dx} = \frac{4}{1} = 4$ For the second derivative, differentiate $\frac{dy}{dx} = \frac{x^2}{y^2}$ using the quotient rule: $\frac{d^2y}{dx^2} = \frac{(y^2)(2x) - (x^2)(2y\frac{dy}{dx})}{y^4} = \frac{2xy^2 - 2x^2y\frac{dy}{dx}}{y^4}$ Factoring: $\frac{d^2y}{dx^2} = \frac{2xy(y - x\frac{dy}{dx})}{y^4} = \frac{2x(y - x\frac{dy}{dx})}{y^3}$ Substituting our values at $(2,1)$ with $\frac{dy}{dx} = 4$: $\frac{d^2y}{dx^2} = \frac{2(2)(1 - 2(4))}{1^3} = \frac{4(1-8)}{1} = 4(-7) = -28$ The answer is (B) -28. Choice (A) 4 is just the first derivative value. Choice (C) -12 likely comes from arithmetic errors in the quotient rule application. Choice (D) 28 represents a sign error—forgetting that $1-8 = -7$, not $+7$. Study tip: In second derivative problems with implicit differentiation, organize your work carefully and double-check arithmetic. The quotient rule creates multiple opportunities for sign and calculation errors.
4. What is the equation of the tangent line to the curve defined by $y^2 + \cos(xy) = x^2$ at the point $(1, 0)$?
   Answer: $x = 1$ First, we find the slope of the tangent line by differentiating the equation implicitly with respect to $x$. We must use the chain rule for the $\cos(xy)$ term, which also requires the product rule for its argument $xy$. $\frac{d}{dx}(y^2 + \cos(xy)) = \frac{d}{dx}(x^2)$$2y\frac{dy}{dx} - \sin(xy) \cdot \left(1 \cdot y + x \frac{dy}{dx}\right) = 2x$ Distribute the $-\sin(xy)$ term: $2y\frac{dy}{dx} - y\sin(xy) - x\sin(xy)\frac{dy}{dx} = 2x$ Group the terms with $\frac{dy}{dx}$ and solve for it: $\frac{dy}{dx}(2y - x\sin(xy)) = 2x + y\sin(xy)$$\frac{dy}{dx} = \frac{2x + y\sin(xy)}{2y - x\sin(xy)}$ Now, evaluate the slope at the point $(1, 0)$: $m = \frac{2(1) + 0 \cdot \sin(0)}{2(0) - 1 \cdot \sin(0)} = \frac{2}{0}$ The slope is undefined, which indicates a vertical tangent line. The equation of a vertical line passing through the point $(1, 0)$ is $x = 1$. Distractor Rationale: (A) This line has a slope of $2$. This slope would be found if the derivative was mistakenly calculated as $\frac{dy}{dx} = 2x$. (B) This is the equation for a horizontal tangent line, which would occur if the slope was $0$. (D) This line has a slope of $-2$, resulting from potential sign errors during differentiation or algebraic manipulation.
5. For the curve $x^2 + kxy + y^2 = 1$, the slope of the tangent line at the point $(1, 0)$ is $-2$. What is the value of the constant $k$?
   Answer: $1$ When you encounter a curve defined implicitly and need to find the slope at a specific point, implicit differentiation is your go-to tool. This technique allows you to find $\frac{dy}{dx}$ without solving explicitly for $y$. Starting with $x^2 + kxy + y^2 = 1$, differentiate both sides with respect to $x$. The left side gives us: $2x + k(y + x\frac{dy}{dx}) + 2y\frac{dy}{dx} = 0$. Notice how the product rule applies to the $kxy$ term, and the chain rule gives us $\frac{dy}{dx}$ terms for $y^2$. Solving for $\frac{dy}{dx}$: $2x + ky + kx\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$. Collecting terms with $\frac{dy}{dx}$: $\frac{dy}{dx}(kx + 2y) = -2x - ky$. Therefore: $\frac{dy}{dx} = \frac{-2x - ky}{kx + 2y}$. At the point $(1, 0)$, this becomes: $\frac{dy}{dx} = \frac{-2(1) - k(0)}{k(1) + 2(0)} = \frac{-2}{k}$. Since we're told the slope equals $-2$, we have: $\frac{-2}{k} = -2$, which gives us $k = 1$. Choice D is correct. Choice A gives slope 1, choice B gives slope 2, and choice C gives slope $-1$ when substituted back into our formula. Remember: with implicit differentiation problems, always substitute your point coordinates into the derivative formula you derived, not into the original equation. This approach works for any implicit curve where you need tangent line information.
6. A student is asked to find $\frac{dy}{dx}$ for the curve $x^2y = \sin(y)$. The student's work is shown below: Line 1: $\frac{d}{dx}(x^2y) = \frac{d}{dx}(\sin(y))$ Line 2: $2xy = \cos(y) \frac{dy}{dx}$ Line 3: $\frac{dy}{dx} = \frac{2xy}{\cos(y)}$ In which line did the student first make a mistake?
   Answer: Line 2, the product rule was not applied correctly. When you encounter implicit differentiation problems like this one, you need to carefully apply differentiation rules to both sides of the equation, remembering that $y$ is a function of $x$. Let's trace through the student's work. Line 1 correctly sets up the problem by differentiating both sides of $x^2y = \sin(y)$. Line 3 correctly isolates $\frac{dy}{dx}$ through algebraic manipulation. However, Line 2 contains a critical error. The mistake occurs in differentiating the left side $x^2y$. Since this is a product of two functions ($x^2$ and $y$), you must use the product rule: $\frac{d}{dx}(uv) = u'v + uv'$. The correct differentiation should be: $\frac{d}{dx}(x^2y) = \frac{d}{dx}(x^2) \cdot y + x^2 \cdot \frac{dy}{dx} = 2xy + x^2\frac{dy}{dx}$ The student wrote $2xy$, which only accounts for the first term of the product rule and ignores the second term $x^2\frac{dy}{dx}$. Answer choice A is incorrect because the initial setup is proper. Answer choice B is wrong since the algebraic manipulation in Line 3 is mathematically sound. Answer choice C misidentifies the error—the chain rule was correctly applied to $\sin(y)$ on the right side, giving $\cos(y)\frac{dy}{dx}$. Answer choice D correctly identifies that the product rule was improperly applied. Study tip: In implicit differentiation, always check if you need the product rule when differentiating terms with both $x$ and $y$. Write out both terms of the product rule explicitly to avoid missing one.
7. Consider the curve defined by the equation $x^2 + y^2 = 25$. What is the value of the second derivative, $\frac{d^2y}{dx^2}$, at the point $(3, 4)$?
   Answer: $-\frac{25}{64}$ First, differentiate the equation implicitly with respect to $x$: $2x + 2y \frac{dy}{dx} = 0$, which gives $\frac{dy}{dx} = -\frac{x}{y}$. To find the second derivative, differentiate this expression using the quotient rule: $\frac{d^2y}{dx^2} = -\frac{(1)y - x(\frac{dy}{dx})}{y^2}$. Substitute $\frac{dy}{dx} = -\frac{x}{y}$ into this equation: $\frac{d^2y}{dx^2} = -\frac{y - x(-\frac{x}{y})}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2} = -\frac{y^2 + x^2}{y^3}$. From the original equation, we know $x^2 + y^2 = 25$. So, $\frac{d^2y}{dx^2} = -\frac{25}{y^3}$. At the point $(3, 4)$, the value is $-\frac{25}{4^3} = -\frac{25}{64}$.
8. Consider the curve defined by the equation $x^2 + y^2 = 25$. What is the value of the second derivative, $d^2y/dx^2$, at the point $(3, 4)$?
   Answer: $-25/64$ First, differentiate the equation implicitly with respect to $x$: $2x + 2y \frac{dy}{dx} = 0$. Solving for $\frac{dy}{dx}$ gives $\frac{dy}{dx} = -\frac{x}{y}$. Next, differentiate this expression with respect to $x$ using the quotient rule to find the second derivative: $\frac{d^2y}{dx^2} = \frac{-\frac{d}{dx}(x) \cdot y - (-x) \cdot \frac{d}{dx}(y)}{y^2} = \frac{-1 \cdot y + x \cdot \frac{dy}{dx}}{y^2}$ Substitute the expression for $\frac{dy}{dx}$: $\frac{d^2y}{dx^2} = \frac{-y + x(-\frac{x}{y})}{y^2} = \frac{-y - \frac{x^2}{y}}{y^2} = \frac{-y^2 - x^2}{y^3}$ From the original equation, we know $x^2 + y^2 = 25$, so we can substitute this into the numerator: $\frac{d^2y}{dx^2} = -\frac{(x^2 + y^2)}{y^3} = -\frac{25}{y^3}$ Finally, evaluate at the point $(3, 4)$: $\frac{d^2y}{dx^2} = -\frac{25}{4^3} = -\frac{25}{64}$ Distractor Rationale: (A) This is the value of the first derivative $\frac{dy}{dx}$ at $(-3, 4)$, not the second derivative at $(3,4)$. (C) This results from a sign error in the quotient rule. (D) This results from an algebraic error in simplification, calculating the denominator as $y^2$ instead of $y^3$.
9. If $x^2 + y^2 = \sin(x+y)$, find the slope of the tangent line at the origin $(0,0)$.
   Answer: $-1$ When you encounter an equation relating x and y that you need to differentiate, you're dealing with implicit differentiation. This technique is essential when you can't easily solve for y in terms of x. Starting with $x^2 + y^2 = \sin(x+y)$, differentiate both sides with respect to x. The left side gives you $2x + 2y\frac{dy}{dx}$ (remember the chain rule for the $y^2$ term). The right side becomes $\cos(x+y) \cdot \left(1 + \frac{dy}{dx}\right)$ using the chain rule on the composite function $\sin(x+y)$. Setting them equal: $2x + 2y\frac{dy}{dx} = \cos(x+y)\left(1 + \frac{dy}{dx}\right)$ At the origin $(0,0)$, this simplifies to: $0 + 0 = \cos(0)\left(1 + \frac{dy}{dx}\right) = 1 + \frac{dy}{dx}$ Solving for the slope: $\frac{dy}{dx} = -1$ Answer A ($1$) represents the positive version of the correct answer—you might get this if you made a sign error when moving terms to isolate $\frac{dy}{dx}$. Answer C ($0$) could result from incorrectly thinking the derivative at the origin is zero because both coordinates are zero. Answer D (undefined) would occur if you mistakenly thought the denominator of your slope calculation was zero. Remember: with implicit differentiation problems, always substitute the given point after finding the general derivative formula, and be extra careful with signs when isolating $\frac{dy}{dx}$.
10. For the curve defined by $x^3 + y^3 = 1$, which of the following is a simplified expression for $\frac{d^2y}{dx^2}$?
    Answer: $-\frac{2x}{y^5}$ First, find $\frac{dy}{dx}$. Differentiating $x^3 + y^3 = 1$ gives $3x^2 + 3y^2 \frac{dy}{dx} = 0$, so $\frac{dy}{dx} = -\frac{x^2}{y^2}$. Next, differentiate this expression using the quotient rule to find $\frac{d^2y}{dx^2}$: $\frac{d^2y}{dx^2} = -\frac{2x(y^2) - x^2(2y \frac{dy}{dx})}{(y^2)^2} = -\frac{2xy^2 - 2x^2y \frac{dy}{dx}}{y^4}$. Substitute the expression for $\frac{dy}{dx}$: $\frac{d^2y}{dx^2} = -\frac{2xy^2 - 2x^2y(-\frac{x^2}{y^2})}{y^4} = -\frac{2xy^2 + \frac{2x^4}{y}}{y^4}$. Multiply the numerator and denominator by $y$ to simplify: $\frac{d^2y}{dx^2} = -\frac{2xy^3 + 2x^4}{y^5} = -\frac{2x(y^3 + x^3)}{y^5}$. Since the original equation is $x^3 + y^3 = 1$, we can substitute this in: $\frac{d^2y}{dx^2} = -\frac{2x(1)}{y^5} = -\frac{2x}{y^5}$.
11. For the curve given by $\sqrt{x} + \sqrt{y} = 4$, the tangent line at a certain point is $y = -3x + 12$. What is the $x$-coordinate of this point of tangency?
    Answer: $1$ This problem tests implicit differentiation and the relationship between a curve's derivative and its tangent line slope. When you have an implicitly defined curve and a given tangent line, you can use the fact that the derivative at the point of tangency equals the tangent line's slope. Starting with $\sqrt{x} + \sqrt{y} = 4$, we find the derivative using implicit differentiation. Taking the derivative of both sides: $\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = 0$ Solving for $\frac{dy}{dx}$: $\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}$ Since the tangent line is $y = -3x + 12$, we know the slope is $-3$ at the point of tangency. Therefore: $-\frac{\sqrt{y}}{\sqrt{x}} = -3$, which gives us $\frac{\sqrt{y}}{\sqrt{x}} = 3$, or $\sqrt{y} = 3\sqrt{x}$. Substituting this into the original equation: $\sqrt{x} + 3\sqrt{x} = 4$, so $4\sqrt{x} = 4$, giving us $\sqrt{x} = 1$, and therefore $x = 1$. Choice B is correct. Choice A ($x = 9$) would make $\sqrt{x} = 3$, requiring $\sqrt{y} = 1$ from the original equation, but this gives a slope of $-\frac{1}{3}$, not $-3$. Choice C ($x = 4$) makes $\sqrt{x} = 2$, requiring $\sqrt{y} = 2$, giving slope $-1$. Choice D ($x = 3$) doesn't even satisfy the original curve equation since $\sqrt{3} + \sqrt{y} = 4$ would require $y = (4-\sqrt{3})^2$, which doesn't yield the correct slope. Remember: when finding points of tangency, both the original equation and the derivative condition must be satisfied simultaneously.
12. What is the slope of the tangent line to the curve $e^y + y = x^2$ at the point $(1, 0)$?
    Answer: $1$ When you encounter an equation like $e^y + y = x^2$ where both variables are mixed together, you need implicit differentiation to find the slope of the tangent line. First, let's verify that $(1, 0)$ is actually on the curve: $e^0 + 0 = 1 + 0 = 1 = 1^2$ ✓ Now differentiate both sides with respect to $x$. The left side requires the chain rule since $y$ is a function of $x$: $\frac{d}{dx}[e^y + y] = \frac{d}{dx}[x^2]$ $e^y \cdot \frac{dy}{dx} + \frac{dy}{dx} = 2x$ Factor out $\frac{dy}{dx}$: $(e^y + 1)\frac{dy}{dx} = 2x$ Solve for the slope: $\frac{dy}{dx} = \frac{2x}{e^y + 1}$ At point $(1, 0)$: $\frac{dy}{dx} = \frac{2(1)}{e^0 + 1} = \frac{2}{1 + 1} = \frac{2}{2} = 1$ The correct answer is C) $1$. Looking at the wrong answers: A) $\frac{1}{2}$ likely comes from incorrectly getting $\frac{1}{2x}$ instead of $\frac{2x}{e^y + 1}$. B) $2$ suggests forgetting to add the $1$ in the denominator, giving $\frac{2}{1}$. D) $-1$ might result from a sign error during differentiation. Study tip: With implicit differentiation problems, always verify your point lies on the curve first, then remember that every term containing $y$ needs $\frac{dy}{dx}$ multiplied to it by the chain rule. Factor out $\frac{dy}{dx}$ and substitute your specific point values last.
13. A curve is defined by $x^3 + by^3 = 1$, where $b$ is a non-zero constant. If the point $(x_0, y_0)$ is on the curve, which of the following expressions represents the slope of the tangent line at that point in terms of $x_0$ and $y_0$ only?
    Answer: $-x_0^2 y_0 / (1 - x_0^3)$ First, we find the derivative $\frac{dy}{dx}$ by implicit differentiation: $\frac{d}{dx}(x^3 + by^3) = \frac{d}{dx}(1)$$3x^2 + b(3y^2) \frac{dy}{dx} = 0$ Solving for $\frac{dy}{dx}$ gives: $3by^2 \frac{dy}{dx} = -3x^2 \implies \frac{dy}{dx} = -\frac{x^2}{by^2}$ This gives the slope at $(x_0, y_0)$ as $m = -\frac{x_0^2}{by_0^2}$. However, the question asks for the slope in terms of $x_0$ and $y_0$ only, so we must eliminate the constant $b$. Since the point $(x_0, y_0)$ is on the curve, it satisfies the original equation: $x_0^3 + by_0^3 = 1$. We can solve this equation for $b$: $by_0^3 = 1 - x_0^3 \implies b = \frac{1 - x_0^3}{y_0^3}$ Now substitute this expression for $b$ into the slope formula: $m = -\frac{x_0^2}{(\frac{1 - x_0^3}{y_0^3})y_0^2} = -\frac{x_0^2}{\frac{1 - x_0^3}{y_0}} = -\frac{x_0^2 y_0}{1 - x_0^3}$ Distractor Rationale: (A) This is the correct derivative but fails to complete the second step of the problem, which is to eliminate the constant $b$. (B) This results from a sign error in the initial differentiation, leading to a positive slope expression. (D) This would result from an algebraic error when solving for $b$, for instance, $by_0^3 = 1 + x_0^3$.
14. For the curve defined by $\ln(y) + x^2y = 4$, which of the following is an expression for $\frac{dy}{dx}$?
    Answer: $\frac{-2xy^2}{1 + x^2y}$ Differentiate the equation implicitly with respect to $x$. For $\ln(y)$, the derivative is $\frac{1}{y} \frac{dy}{dx}$ by the chain rule. For $x^2y$, use the product rule: $2xy + x^2 \frac{dy}{dx}$. The derivative of the constant 4 is 0. So, the differentiated equation is $\frac{1}{y}\frac{dy}{dx} + 2xy + x^2\frac{dy}{dx} = 0$. Now, solve for $\frac{dy}{dx}$: $\frac{dy}{dx}(\frac{1}{y} + x^2) = -2xy$. This gives $\frac{dy}{dx} = \frac{-2xy}{\frac{1}{y} + x^2}$. To simplify, multiply the numerator and denominator by $y$: $\frac{dy}{dx} = \frac{-2xy^2}{y(\frac{1}{y} + x^2)} = \frac{-2xy^2}{1 + x^2y}$.
15. For the curve defined by the equation $e^{xy} = x - y$, what is the slope of the tangent line at the point $(1, 0)$?
    Answer: $1/2$ To find the slope, we differentiate the equation implicitly with respect to $x$. This requires the chain rule on the left side and the product rule for the exponent $xy$. $\frac{d}{dx}(e^{xy}) = \frac{d}{dx}(x - y)$$e^{xy} \cdot \frac{d}{dx}(xy) = 1 - \frac{dy}{dx}$$e^{xy} \cdot \left(1 \cdot y + x \frac{dy}{dx}\right) = 1 - \frac{dy}{dx}$ Distribute the $e^{xy}$ term: $y e^{xy} + x e^{xy} \frac{dy}{dx} = 1 - \frac{dy}{dx}$ Now, group all terms with $\frac{dy}{dx}$ on one side and solve: $x e^{xy} \frac{dy}{dx} + \frac{dy}{dx} = 1 - y e^{xy}$$\frac{dy}{dx} (x e^{xy} + 1) = 1 - y e^{xy}$$\frac{dy}{dx} = \frac{1 - y e^{xy}}{x e^{xy} + 1}$ Finally, substitute the coordinates of the point $(1, 0)$ into this expression: $m = \frac{1 - 0 \cdot e^{1 \cdot 0}}{1 \cdot e^{1 \cdot 0} + 1} = \frac{1 - 0}{1 \cdot 1 + 1} = \frac{1}{2}$ Distractor Rationale: (A) This results from forgetting to apply the chain rule to the $-y$ term on the right-hand side, treating its derivative as $0$ instead of $\frac{dy}{dx}$. (B) This results from forgetting the product rule for the term $xy$ in the exponent, treating its derivative as just $y$ or $x$. (C) This results from a sign error when isolating the $\frac{dy}{dx}$ term, for example, getting $\frac{y e^{xy} - 1}{x e^{xy} + 1}$.
16. For the curve defined by the relation $x^y = y^x$, what is the value of $\frac{dy}{dx}$ at the point $(4, 4)$?
    Answer: $1$ When you encounter an implicitly defined relation like $x^y = y^x$, you need implicit differentiation to find $\frac{dy}{dx}$. This technique treats $y$ as a function of $x$ and applies the chain rule whenever you differentiate terms involving $y$. Start by taking the natural logarithm of both sides: $\ln(x^y) = \ln(y^x)$, which simplifies to $y\ln(x) = x\ln(y)$. Now differentiate both sides with respect to $x$: Left side: $\frac{d}{dx}[y\ln(x)] = y \cdot \frac{1}{x} + \ln(x) \cdot \frac{dy}{dx}$ Right side: $\frac{d}{dx}[x\ln(y)] = \ln(y) + x \cdot \frac{1}{y} \cdot \frac{dy}{dx}$ Setting them equal: $\frac{y}{x} + \ln(x)\frac{dy}{dx} = \ln(y) + \frac{x}{y}\frac{dy}{dx}$ Solve for $\frac{dy}{dx}$: $\frac{dy}{dx}[\ln(x) - \frac{x}{y}] = \ln(y) - \frac{y}{x}$ Therefore: $\frac{dy}{dx} = \frac{\ln(y) - \frac{y}{x}}{\ln(x) - \frac{x}{y}}$ At point $(4,4)$: $\frac{dy}{dx} = \frac{\ln(4) - 1}{\ln(4) - 1} = 1$ Answer (A) is wrong because the derivative exists at this point. Answer (B) is incorrect because the numerator and denominator are equal but non-zero. Answer (C) results from forgetting to subtract the fraction terms or making algebraic errors in the simplification. Remember: when working with exponential equations involving both variables, taking logarithms first often simplifies the implicit differentiation process significantly.
17. The line $y = -x + k$ is tangent to the folium of Descartes, $x^3 + y^3 - 6xy = 0$, at a point in the first quadrant. What is the value of $k$?
    Answer: $6$ The slope of the tangent line $y = -x + k$ is -1. We need to find the point on the curve where $\frac{dy}{dx} = -1$. Differentiating the curve's equation: $3x^2 + 3y^2\frac{dy}{dx} - (6y + 6x\frac{dy}{dx}) = 0$. Solving for $\frac{dy}{dx}$ yields $\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}$. Setting the slope to -1: $\frac{2y - x^2}{y^2 - 2x} = -1$, which implies $2y - x^2 = -y^2 + 2x$, or $y^2 + 2y = x^2 + 2x$. Completing the square gives $(y+1)^2 = (x+1)^2$, so $y+1 = \pm(x+1)$. This leads to two cases: $y=x$ or $y=-x-2$. Substituting $y=x$ into the original equation gives $2x^3 - 6x^2 = 0$, with solutions $x=0$ (the origin, a singular point) and $x=3$. The point in the first quadrant is $(3, 3)$. Substituting $y=-x-2$ yields $-8=0$, which is impossible. So the point of tangency is $(3, 3)$. The tangent line $y = -x + k$ must pass through this point: $3 = -3 + k$, which gives $k = 6$.
18. For the curve defined by $x^2 + 2xy + 3y^2 = 6$, there are two points where the tangent line is horizontal. What is the product of the $y$-coordinates of these two points?
    Answer: $-3$ First, find the derivative $\frac{dy}{dx}$ using implicit differentiation: $2x + (2y + 2x\frac{dy}{dx}) + 6y\frac{dy}{dx} = 0$. Grouping terms with $\frac{dy}{dx}$ gives $\frac{dy}{dx}(2x + 6y) = -2x - 2y$, so $\frac{dy}{dx} = -\frac{x+y}{x+3y}$. A horizontal tangent occurs when $\frac{dy}{dx} = 0$, which means the numerator must be zero: $-(x+y) = 0$, or $y = -x$. Substitute this into the original equation: $x^2 + 2x(-x) + 3(-x)^2 = 6$, which simplifies to $x^2 - 2x^2 + 3x^2 = 6$, or $2x^2 = 6$. Thus, $x^2 = 3$, so $x = \sqrt{3}$ and $x = -\sqrt{3}$. The corresponding $y$-coordinates are $y = -\sqrt{3}$ and $y = \sqrt{3}$. The product of these $y$-coordinates is $(-\sqrt{3})(\sqrt{3}) = -3$.
19. The curve given by $\arctan(\frac{y}{x}) = \ln(\sqrt{x^2+y^2})$ describes a logarithmic spiral. What is the slope of the tangent line to this curve at the point $(1, 0)$?
    Answer: $1$ First, simplify the equation: $\arctan(\frac{y}{x}) = \frac{1}{2}\ln(x^2+y^2)$. Differentiate both sides with respect to $x$. The derivative of the left side is $\frac{1}{1+(y/x)^2} \cdot \frac{x(dy/dx) - y}{x^2} = \frac{x^2}{x^2+y^2} \cdot \frac{x(dy/dx) - y}{x^2} = \frac{x(dy/dx) - y}{x^2+y^2}$. The derivative of the right side is $\frac{1}{2} \cdot \frac{1}{x^2+y^2} \cdot (2x + 2y\frac{dy}{dx}) = \frac{x+y(dy/dx)}{x^2+y^2}$. Setting the derivatives equal and multiplying by $x^2+y^2$ gives $x\frac{dy}{dx} - y = x + y\frac{dy}{dx}$. Solving for $\frac{dy}{dx}$ yields $\frac{dy}{dx}(x-y) = x+y$, so $\frac{dy}{dx} = \frac{x+y}{x-y}$. Evaluating at the point $(1, 0)$, the slope is $\frac{1+0}{1-0} = 1$.
20. The normal line to the circle $x^2+y^2=25$ at the point $(3,4)$ intersects the $x$-axis at which point?
    Answer: $(0, 0)$ When finding where a normal line intersects an axis, you need to understand the geometric relationship between circles, tangent lines, and normal lines. For any circle centered at the origin, the normal line at any point passes through both that point and the center. Let's work through this step-by-step. First, verify that $(3,4)$ is actually on the circle: $3^2 + 4^2 = 9 + 16 = 25$ ✓ For the circle $x^2 + y^2 = 25$ centered at $(0,0)$, the normal line at point $(3,4)$ must pass through the center $(0,0)$ and the point $(3,4)$. Since the normal line already passes through $(0,0)$, it intersects the x-axis at the origin. You can verify this algebraically by finding the normal line's equation. The slope from $(0,0)$ to $(3,4)$ is $\frac{4-0}{3-0} = \frac{4}{3}$. Using point-slope form with point $(0,0)$: $y = \frac{4}{3}x$. Setting $y = 0$: $0 = \frac{4}{3}x$, so $x = 0$. Now for the wrong answers: Choice A, $(\frac{25}{3}, 0)$, might tempt you if you incorrectly try to use the radius value of 25 in some calculation. Choice C, $(3,0)$, could seem appealing if you mistakenly think the x-coordinate of the point on the circle determines where the normal hits the x-axis. Choice D, $(4,0)$, similarly uses the y-coordinate incorrectly. Key insight: For any circle centered at the origin, the normal line at any point always passes through the center, making problems like this much simpler than they initially appear.

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