8.1 Concept of implicit differentiation

Implicit differentiation is a powerful tool for finding derivatives of functions that aren't explicitly defined. It's especially handy when dealing with equations where x and y terms are mixed together, like in circles or other complex shapes.

This technique uses the chain rule to differentiate both sides of an equation with respect to x, treating y as a function of x. It allows us to find derivatives and tangent lines for curves that would be tricky to work with otherwise.

Implicit Differentiation

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Concept of implicit differentiation

• Technique for finding derivatives of functions not explicitly defined as $y = f(x)$
• Useful when both $x$ and $y$ terms are on the same side of the equation ($x^2 + y^2 = 25$)
• Necessary when solving for $y$ in terms of $x$ is difficult or impossible
• Applies when a function is defined by a relationship between $x$ and $y$ rather than an explicit formula

Chain rule for implicit functions

• Differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$
• When differentiating a term involving $y$, apply the chain rule: $\frac{d}{dx}(y) = \frac{dy}{dx}$
• Implicitly differentiate $x^2 + y^2 = 25$:
  1. $\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)$
  2. $2x + 2y\frac{dy}{dx} = 0$

Derivatives of implicit functions

• After implicit differentiation, solve for $\frac{dy}{dx}$ to find the derivative of the implicitly defined function
• Find $\frac{dy}{dx}$ for the implicitly defined function $x^2 + y^2 = 25$:
  1. $2x + 2y\frac{dy}{dx} = 0$ (from previous example)
  2. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = -\frac{x}{y}$

Tangent lines to implicit curves

• To find the equation of a tangent line at a point $(a, b)$ on an implicitly defined curve:
  1. Find $\frac{dy}{dx}$ using implicit differentiation
  2. Evaluate $\frac{dy}{dx}$ at the point $(a, b)$ to find the slope of the tangent line
  3. Use the point-slope form of a line: $y - b = m(x - a)$, where $m$ is the slope
• Find the equation of the tangent line to the curve $x^2 + y^2 = 25$ at the point $(3, 4)$:
  1. $\frac{dy}{dx} = -\frac{x}{y}$ (from previous example)
  2. Evaluate $\frac{dy}{dx}$ at $(3, 4)$: $\frac{dy}{dx}|_{(3, 4)} = -\frac{3}{4}$
  3. Use the point-slope form: $y - 4 = -\frac{3}{4}(x - 3)$