12-5-24 Implicit Differentiation Intro (virtual)

Front

What is Implicit Differentiation?

Back

Implicit differentiation is a technique used to find the derivative of a function defined implicitly by an equation involving both x and y, without solving for y explicitly.

Front

Back

Differentiate both sides with respect to x, treating y as a function of x: \(3x^2 + 3y^2 \frac{dy}{dx} = 0\). Solve for \(\frac{dy}{dx}\) to get \(\frac{dy}{dx} = -\frac{x^2}{y^2}\).

Front

Back

Differentiate both sides: \(2y \frac{dy}{dx} = 10\). Solve for \(\frac{dy}{dx}\) to get \(\frac{dy}{dx} = \frac{5}{y}\).

Front

Back

Differentiate both sides: \(4x - 6y \frac{dy}{dx} = 0\). Solve for \(\frac{dy}{dx}\) to get \(\frac{dy}{dx} = \frac{2x}{3y}\).

Front

Back

Differentiate both sides: \(y + x \frac{dy}{dx} + 2y \frac{dy}{dx} = 0\). Solve for \(\frac{dy}{dx}\) to get \(\frac{dy}{dx} = -\frac{y}{x + 2y}\).

Front

Back

Differentiate both sides: \(2x + y + x \frac{dy}{dx} + 3y^2 \frac{dy}{dx} = 0\). Solve for \(\frac{dy}{dx}\) to get \(\frac{dy}{dx} = -\frac{2x + y}{x + 3y^2}\).

Front

What is the Chain Rule in the context of implicit differentiation?

Back

The Chain Rule states that if a variable depends on another variable, the derivative of the outer function is multiplied by the derivative of the inner function. In implicit differentiation, it is used when differentiating terms involving y.

Front

What is the significance of treating y as a function of x in implicit differentiation?

Back

Treating y as a function of x allows us to apply the rules of differentiation to find the slope of the tangent line to the curve defined by the implicit equation.

Front

What is the first step in implicit differentiation?

Back

The first step is to differentiate both sides of the equation with respect to x, applying the product rule and chain rule as necessary.

Front

What is the product rule in differentiation?

Back

The product rule states that if you have two functions u(x) and v(x), then the derivative of their product is given by: \(\frac{d}{dx}[u v] = u'v + uv'\).