Implicit Differentiation: Second Derivatives & Critical Points

Differentiate both sides with respect to x: 2x + 2y(dy/dx) = 0. Solve for dy/dx by isolating the term containing dy/dx: 2y(dy/dx) = -2x. Divide both sides by 2y to obtain dy/dx = -2x / 2y = -x/y.

Differentiate both sides: use product rule on x²y → 2xy + x²(dy/dx); for y³ → 3y²(dy/dx); right side 4. The equation becomes 2xy + x²(dy/dx) + 3y²(dy/dx) = 4. Group dy/dx terms: (x² + 3y²)(dy/dx) = 4 - 2xy. Thus dy/dx = (4 - 2xy)/(x² + 3y²).

Start with xy = 8. Differentiate implicitly: y + x(dy/dx) = 0. Solve for dy/dx: x(dy/dx) = -y → dy/dx = -y/x. At the point (2,4), substitute x = 2 and y = 4 to get dy/dx = -4/2 = -2.

Take natural log first or differentiate directly. Differentiate eˣʸ = y: left side eˣʸ · (y + x dy/dx) = dy/dx (chain rule on exponent). So eˣʸ(y + x dy/dx) = dy/dx. Since eˣʸ = y, substitute: y(y + x dy/dx) = dy/dx. Expand: y² + xy dy/dx = dy/dx. Bring terms with dy/dx to one side: xy dy/dx - dy/dx = -y². Factor: dy/dx (xy - 1) = -y². Thus dy/dx = -y²/(xy - 1) = y²/(1 - xy).

To find where the tangent line is horizontal, we must find the points where the derivative dy/dx equals zero. First, we perform implicit differentiation on the equation x³ + y³ = 6xy with respect to x. The derivative of x³ is 3x², and the derivative of y³ is 3y²(dy/dx). For the term 6xy, we use the product rule, yielding 6(1*y + x(dy/dx)). This gives the equation 3x² + 3y²(dy/dx) = 6y + 6x(dy/dx). We then divide by 3 and group the dy/dx terms to solve for the derivative: dy/dx(y² - 2x) = 2y - x², which simplifies to dy/dx = (2y - x²) / (y² - 2x). For the tangent to be horizontal, the numerator must be zero (and the denominator non-zero), so we just have to check which of the choices satisfy the equation 2y - x²= 0. Only choice C satisfies this requirement.

Differentiate: 2x + 8y dy/dx = 0 → dy/dx = -2x/(8y) = -x/(4y). At (2√3, 1), slope = -(2√3)/(4·1) = -√3/2. Point-slope form: y - 1 = (-√3/2)(x - 2√3). Simplify: y - 1 = (-√3/2)x + √3 · √3 = (-√3/2)x + 3. Thus y = (-√3/2)x + 4.

Differentiate both sides to obtain cos(x + y) · (1 + dy/dx) = dy/dx. We expand the left side to get cos(x + y) + cos(x + y) dy/dx = dy/dx. Grouping dy/dx terms, gives cos(x + y) = dy/dx - cos(x + y) dy/dx = dy/dx (1 - cos(x + y)). Thus dy/dx = cos(x + y)/(1 - cos(x + y)).

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From question 1, dy/dx = -x/y. Differentiate again implicitly (quotient or product): d/dx [-x y⁻¹] = -1 · y⁻¹ - x (-1) y⁻² dy/dx = -1/y + (x/y²)(dy/dx). Substitute dy/dx = -x/y: -1/y + (x/y²)(-x/y) = -1/y - x²/y³. Common denominator y³: (-y² - x²)/y³ = -(x² + y²)/y³ = -4/y³.

Differentiate: (dy/dx) ln x + y (1/x) = 1 - dy/dx. Bring dy/dx terms together: dy/dx ln x + dy/dx = 1 - y/x. Factor dy/dx (ln x + 1) = 1 - y/x. At (e,1): dy/dx (1 + 1) = 1 - 1/e → 2 dy/dx = (e - 1)/e → dy/dx = (e - 1)/(2e).

Differentiate: 3x² + 3y² dy/dx - 9(y + x dy/dx) = 0 → 3x² + 3y² dy/dx - 9y - 9x dy/dx = 0. Group: dy/dx (3y² - 9x) = 9y - 3x² → dy/dx = (3y - x²)/(y² - 3x). At (0,0), denominator = 0 - 0 = 0 and numerator = 0, giving 0/0 indeterminate, but limit analysis shows vertical tangent; slope undefined.

Differentiate: -sin(xy) (y + x dy/dx) = dy/dx. Expand: -y sin(xy) - x sin(xy) dy/dx = dy/dx. Bring terms: -y sin(xy) = dy/dx + x sin(xy) dy/dx = dy/dx (1 + x sin(xy)). Thus dy/dx = -y sin(xy)/(1 + x sin(xy)).

We start by differentiating both sides of the equation with respect to x. On the left side, we use the chain rule on (x² + y²)² to get 2(x² + y²) multiplied by the derivative of the inner term, which is (2x + 2y dy/dx). On the right side, we use the product rule on 4xy to get 4(y + x dy/dx). Setting these equal gives us the equation 2(x² + y²)(2x + 2y dy/dx) = 4y + 4x dy/dx. We can now substitute x = 1 and y = 1 into the equation to solve for dy/dx. This gives 2(1² + 1²)(2(1) + 2(1) dy/dx) = 4(1) + 4(1) dy/dx, which simplifies to 4(2 + 2 dy/dx) = 4 + 4 dy/dx. Expanding the left side results in 8 + 8 dy/dx = 4 + 4 dy/dx. Subtracting 4 dy/dx from both sides yields 4 dy/dx = -4. Finally, dividing by 4 gives dy/dx = -1.

Standard ellipse. dy/dx = - (x/9) · (16/y) = -16x/(9y). Set = 0 → x = 0, then plug into the ellipse: y = ±4.

First find dy/dx to get 2y + 2x dy/dx + (1/y) dy/dx = 1. At (1,1): 2 + 2 dy/dx + dy/dx = 1 → 3 dy/dx = -1 → dy/dx = -1/3. Then differentiate again implicitly to obtain 2 dy/dx + 2(dy/dx + x d²y/dx²) + (y (d²y/dx²) - (dy/dx)²)/(y²) = 0. Plugging in the values into this equation involving the second derivative, we have 2(-1/3) + 2(-1/3 + d²y/dx²) + (d²y/dx²) - (-1/3)² = 0. This gives d²y/dx² = 13/27.

We start by modeling the ladder, wall, and ground as a right triangle where x is the distance from the wall and y is the height on the wall. The length of the ladder is constant at 10 feet, so by the Pythagorean theorem, we have the equation x² + y² = 100. Next, we differentiate both sides of this equation with respect to time t to relate the rates of change. The derivative of x² is 2x(dx/dt) and the derivative of y² is 2y(dy/dt), while the derivative of the constant 100 is 0. This gives us the equation 2x(dx/dt) + 2y(dy/dt) = 0, which simplifies to x(dx/dt) + y(dy/dt) = 0. We are given that x = 6 feet and the bottom moves away at dx/dt = 2 ft/s. Before we can solve for dy/dt, we must find the current height y. Substituting x = 6 into the original Pythagorean equation, we get 6² + y² = 100, which means 36 + y² = 100. Solving for y, we find y² = 64, so y = 8 feet. Finally, we substitute these values into our differentiated equation: (6)(2) + (8)(dy/dt) = 0. This simplifies to 12 + 8(dy/dt) = 0. Subtracting 12 from both sides gives 8(dy/dt) = -12. Dividing by 8, we get dy/dt = -12/8, which simplifies to -3/2 ft/s.