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Implicit differentiation

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From: 3blue1brown

Implicit differentiation is a technique used in calculus to find the slope of a tangent line to a curve or the rate of change of variables in an equation where one variable is not explicitly defined as a function of the other [01:31:00]. This method is particularly useful for “implicit curves,” which are sets of points (x, y) that satisfy a property written in terms of two variables, x and y [01:52:04].

Finding the Slope of a Tangent Line to a Circle

Consider a circle with radius 5 centered at the origin, defined by the equation x² + y² = 5² [00:00:22]. To find the slope of a tangent line to this circle, for instance at the point (3,4) [00:00:44], a direct application of simple derivatives is not possible because the curve is not the graph of a single function [01:31:00]. X is not an input, and y is not an output; they are interdependent values [01:44:26].

The key idea for finding tangent line slopes is to “zoom in” until the curve looks like its tangent line, then consider a tiny step along the curve [01:07:11]. The y-component of this step is dy, and the x-component is dx, making the desired slope dy/dx [01:17:21].

To apply implicit differentiation:

  1. Take the derivative of both sides of the equation x² + y² = 5² [02:12:12].
    • For x², the derivative is 2x * dx [02:16:32].
    • For y², the derivative is 2y * dy [02:19:50].
    • The derivative of the constant 5² is 0 [02:24:25].
    • This yields the equation: 2x dx + 2y dy = 0 [02:46:17].
  2. Rearrange the equation to find an expression for dy/dx [02:46:17].

At the point (x, y) = (3, 4), the slope dy/dx is -3/4 [02:56:56].

Implicit differentiation is conceptually linked to “related rates” problems [03:20:00].

Example: The Ladder Problem

Imagine a 5-meter ladder leaning against a wall [03:26:34]. If the top of the ladder is 4 meters above the ground and is slipping down at 1 meter per second, the bottom is initially 3 meters from the wall [03:30:19]. The question is, at what rate is the bottom of the ladder moving away from the wall at that moment [03:46:42]?

  1. Define variables: Let y(t) be the distance from the top of the ladder to the ground, and x(t) be the distance from the bottom of the ladder to the wall [04:21:49].
  2. The relationship is given by the Pythagorean theorem: x(t)² + y(t)² = 5² [04:34:01]. This equation holds true at all times [04:43:40].
  3. Take the derivative of both sides with respect to time (t) [05:36:58].
  4. Substitute known values at t=0: y(t) = 4m, x(t) = 3m, and dy/dt = -1 m/s (negative because it’s dropping) [06:45:51].
    • 2(3) (dx/dt) + 2(4) (-1) = 0
    • 6 (dx/dt) - 8 = 0
    • dx/dt = 8/6 = 4/3 meters per second [07:04:15].

In the ladder problem, taking the derivative has a clear meaning: it’s the rate at which the expression changes as time changes [07:32:04]. For the circle, the “tiny nudges dx and dy” are not tied to an external variable like time [07:52:16], which initially feels strange [07:43:08].

Interpretation of Implicit Differentiation

To understand the meaning of taking a derivative of an expression with multiple variables:

  1. Let s = x² + y² [08:03:07]. This s is a function of two variables, x and y [08:08:11]. For points on the circle, s equals 25 [08:16:38].
  2. Taking a derivative of this expression means considering a tiny change dx to x and a tiny change dy to y [08:34:03].
  3. The change in s, denoted ds, is approximately 2x dx + 2y dy [09:21:40]. This formula tells you how much the value of x² + y² changes based on your starting point (x,y) and the tiny step (dx, dy) [09:41:43].
  4. When you restrict yourself to steps along the circle, you are ensuring that the value of s does not change; it remains 25 [10:07:00].
  5. Therefore, ds must be 0 [10:17:34].
  6. Setting the expression 2x dx + 2y dy equal to 0 is the condition for a tiny step to stay on the tangent line of the circle [10:20:23]. For tiny enough steps, the tangent line and the circle are essentially the same [10:40:02].

Generalizing Implicit Differentiation

Implicit differentiation can be applied to any equation relating x and y.

Example: sin(x)y² = x

Consider the equation sin(x)y² = x, which represents a set of u-shaped curves [10:53:23].

  1. Take the derivative of each side with respect to x (implicitly) [11:23:42].
    • For the left side, sin(x)y², apply the product rule: “left d right plus right d left” [11:32:06].
      • sin(x) * (change to y²), which is sin(x) * 2y dy [11:39:10].
      • y² * (change to sin(x)), which is y² * cos(x) dx [11:44:03].
    • For the right side, x, the change is simply dx [11:52:16].
  2. Set the changes of both sides equal to each other to ensure the step stays on the curve [11:59:04].
  3. From here, you can algebraically rearrange to find dy/dx [12:26:17].

Deriving New Derivative Formulas

Implicit differentiation can also be used to derive new derivative formulas.

Example: Derivative of Natural Logarithm ln(x)

To find the derivative of ln(x):

  1. Let y = ln(x) [12:50:00]. This implicitly defines the curve.
  2. Rearrange the equation using the definition of natural logarithm: e^y = x [13:16:16].
  3. Take the derivative of both sides, considering how a tiny step (dx, dy) changes the value of each side [13:31:00].
  4. Equate the changes to ensure the step stays on the curve: e^y dy = dx [13:50:20].
  5. Rearrange to solve for dy/dx: dy/dx = 1 / e^y [13:57:38].
  6. Since e^y is equal to x when on the curve, substitute x for e^y: dy/dx = 1 / x [14:06:17].

This demonstrates that the derivative of ln(x) is 1/x [14:24:20].

Implicit differentiation is a fundamental concept in multivariable calculus that helps understand how functions with multiple inputs change when those inputs are tweaked [14:32:00].

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