Implicit Differentiation with Exponentials and Logarithms

Implicit differentiation relies on the chain rule to handle the fact that y is a function of x. When we differentiate terms involving y, we must apply the chain rule by multiplying by dy/dx to account for the indirect dependence of y on x through the functional relationship defined by the equation.

Differentiating both sides: d/dx[x²e^y] + d/dx[yeˣ] = 0. For x²ey: 2xey + x²ey(dy/dx). For yeˣ: eˣ(dy/dx) + yeˣ. So we have: 2xey + x²ey(dy/dx) + eˣ(dy/dx) + yeˣ = 0. Grouping dy/dx terms: x²ey(dy/dx) + eˣ(dy/dx) = -2xey - yeˣ. So (x²ey + eˣ)(dy/dx) = -(2xey + yeˣ). Therefore: dy/dx = -(2xey + yeˣ)/(x²ey + eˣ).

A critical point on any curve occurs when the first derivative equals zero or is undefined. This is the definition for both explicitly and implicitly defined functions. Critical points correspond to locations where the tangent line is horizontal (dy/dx = 0) or vertical (dy/dx undefined), which is when the curve changes direction in terms of increasing/decreasing behavior.

First verify (1, 1) is on the curve: 1 + 1 + 1 = 3, which is correct. Differentiating: 2x + 2y(dy/dx) + [y + x(dy/dx)] = 0. So 2x + 2y(dy/dx) + y + x(dy/dx) = 0. Collecting dy/dx terms: 2y(dy/dx) + x(dy/dx) = -2x - y. So (2y + x)(dy/dx) = -(2x + y). Therefore: dy/dx = -(2x + y)/(x + 2y). At (1, 1): dy/dx = -(2(1) + 1)/(1 + 2(1)) = -(2 + 1)/(1 + 2) = -3/3 = -1.

In implicit differentiation, we treat y as an implicit function of x. Therefore, when differentiating terms containing y with respect to x, we must apply the chain rule. For the term y², the derivative with respect to x is 2y * (dy/dx), not simply 2y. This is because the chain rule accounts for the fact that y itself changes as x changes. This fundamental principle of implicit differentiation allows us to find derivatives even when we cannot solve explicitly for y in terms of x.

Using the chain rule, d/dx[cos(u)] = -sin(u) · du/dx, where u = xy. So d/dx[cos(xy)] = -sin(xy) · d/dx(xy) = -sin(xy) · (y + x dy/dx) using the product rule.

Differentiating: (3/2)x½ + (3/2)y½(dy/dx) = 0. Multiply both sides by 2/3: x½ + y½(dy/dx) = 0. So y½(dy/dx) = -x½. Therefore: dy/dx = -x½/(y½) = -√(x/y).

An elliptical orbit is defined by an equation where y is not a simple function of x. To find the slope of the path at any point, which represents the instantaneous direction of motion in the xy-plane, we must find dy/dx. Since we cannot easily solve for y, we differentiate the equation x²/a² + y²/b² = 1 implicitly with respect to x to find an expression for dy/dx in terms of x and y.

Differentiating: d/dx[x³y²] + d/dx[2xy³] = 0. For x³y²: 3x²y² + x³·2y(dy/dx) = 3x²y² + 2x³y(dy/dx). For 2xy³: 2[y³ + x·3y²(dy/dx)] = 2y³ + 6xy²(dy/dx). So total: 3x²y² + 2x³y(dy/dx) + 2y³ + 6xy²(dy/dx) = 0. Grouping dy/dx terms: 2x³y(dy/dx) + 6xy²(dy/dx) = -3x²y² - 2y³. So (2x³y + 6xy²)(dy/dx) = -(3x²y² + 2y³). Therefore: dy/dx = -(3x²y² + 2y³)/(2x³y + 6xy²).

Find dy/dx: 3y²(dy/dx) = 2x(5 - x) + x²(-1) = 10x - 3x². So dy/dx = (10x - 3x²)/(3y²). Critical points occur when dy/dx = 0 or undefined. Setting numerator = 0 gives x = 0 or x = 10/3. In the first quadrant, x = 10/3 gives y > 0. Also check y = 0 gives x = 0 or 5, which are undefined slopes. Thus there are 2 critical points.

Differentiating both sides: sec²(x + y)(1 + dy/dx) = 1. Solving gives dy/dx = (1 - sec²(x + y))/sec²(x + y).

The chain rule is important in implicit differentiation because we treat y as a function of x. When differentiating terms containing y, we must multiply by dy/dx to account for this functional relationship.

Differentiating P·Q = K gives dQ/dP = -Q/P. This represents how quantity changes as price changes.

Differentiating both sides and solving gives dy/dx = (2x - y)/(x - 2y).

Implicit differentiation gives dy/dx = -x/y, representing how one concentration changes relative to the other at equilibrium.