Implicit Differentiation Fundamentals

When using implicit differentiation, the first step is always to differentiate both sides of the equation with respect to x. This allows us to find dy/dx even when y cannot be solved for explicitly. The chain rule will be applied to any terms containing y since y is a function of x.

Differentiating both sides with respect to x gives 3x² + 3y²(dy/dx) = 0. Subtract 3x² from both sides: 3y²(dy/dx) = -3x². Divide both sides by 3y²: dy/dx = -x²/y².

Implicit differentiation is used when y cannot be easily solved for in terms of x. Options A, B, and D show y explicitly as a function of x, while option C relates x and y in a way that requires implicit differentiation to find dy/dx.

From x² + y² = 25, differentiating implicitly gives 2x + 2y(dy/dx) = 0. Solving for dy/dx: dy/dx = -x/y. At (3, 4), dy/dx = -3/4.

Differentiating both sides: eˣ + ey(dy/dx) = 0. Solving for dy/dx: dy/dx = -eˣ/ey. At (0, 0), dy/dx = -e⁰/e⁰= -1/1 = -1.

Using the chain rule, d/dx[sin(y)] = cos(y) · dy/dx. This is because y is a function of x, so we must multiply by the derivative of y with respect to x.

To find dy/dx, we differentiate both sides of the equation xy = 6 with respect to x. On the left side, we must use the product rule: d/dx[xy] = (d/dx[x]) * y + x * (d/dx[y]) = 1*y + x*(dy/dx). On the right side, the derivative of the constant 6 is 0. Setting the derivatives equal gives us the equation: y + x(dy/dx) = 0. To solve for dy/dx, we rearrange the terms: x(dy/dx) = -y, which gives dy/dx = -y/x. The question requires the answer to be in terms of x only. From the original equation, we can write y = 6/x. Substituting this into our expression for dy/dx yields dy/dx = -(6/x) / x = -6/x².

When using implicit differentiation, if the expression for dy/dx has a denominator that equals zero at a particular point, then dy/dx is undefined at that point. This can correspond to vertical tangent lines.

Differentiating both sides: d/dx(x²y) + d/dx(y³) = d/dx(12). Using product rule on x²y: 2xy + x²(dy/dx). For y³: 3y²(dy/dx). So we have 2xy + x²(dy/dx) + 3y²(dy/dx) = 0. Grouping dy/dx terms: x²(dy/dx) + 3y²(dy/dx) = -2xy. So (x² + 3y²)(dy/dx) = -2xy. Therefore: dy/dx = -2xy/(x² + 3y²).

We have dy/dx = -x/y. Setting the slope equal to 0: -x/y = 0, which gives x = 0. Substituting x = 0 into x² + y² = 25: 0 + y² = 25, so y = ±5. Therefore, the points are (0, 5) and (0, -5).

Differentiating both sides: 2y(dy/dx) = 4. Dividing both sides by 2y: dy/dx = 4/(2y) = 2/y.

First find dy/dx: 2x + 2y(dy/dx) = 0, so dy/dx = -x/y. Now find d²y/dx² by differentiating dy/dx using the Quotient Rule:d²y/dx² = [(-1)·y - (-x)·(dy/dx)] / y² = [-y + x(dy/dx)] / y². Now, substitute the expression for dy/dx back into this equation to obtain d²y/dx² = [-y + x(-x/y)] / y². Multiply the numerator and denominator by y to get d²y/dx² = [y(-y) + x(-x)] / y³ = [-y² - x²] / y³. Factor out -1 from the numerator gives d²y/dx² = -(x² + y²) / y³. From the original equation, we know x² + y² = 9. Substitute this into the numerator to get d²y/dx² = -9 / y³.

The derivative of ln(u) with respect to x is (1/u) · du/dx, where u = x + y. So d/dx[ln(x + y)] = 1/(x + y) · d/dx(x + y) = 1/(x + y) · (1 + dy/dx).

Differentiating both sides: cos(xy) · d/dx(xy) = 1. Using product rule: cos(xy) · (x(dy/dx) + y) = 1. So cos(xy) · x(dy/dx) + cos(xy) · y = 1. Solving: x cos(xy)(dy/dx) = 1 - y cos(xy). Therefore: dy/dx = (1 - y cos(xy))/(x cos(xy)).

Implicit differentiation is most useful when we cannot or do not want to solve for y explicitly as a function of x. This includes cases where solving would be extremely difficult (like complex algebraic equations) or where multiple y-values correspond to a single x-value (like circles, ellipses, and other implicit relations).