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Implicit Differentiation


Implicit differentiation is a powerful technique in calculus that allows us to find the derivative of a function even when it is not easy or possible to solve the function for one variable in terms of the other. In simple terms, it helps us differentiate a function given in implicit form. An implicit function is an equation where the dependent and independent variables are intertwined and cannot be separated easily.

Understanding built-in functions

Implicit functions involve two or more variables that are related through an equation. For example, consider the equation of a circle:

x² + y² = r²

In this equation, x and y are variables that are connected in such a way that y cannot be easily solved for in terms of x or vice versa. This is in contrast to an explicit function, such as y = x², where y is expressed explicitly as a function of x.

Underlying differentiation process

Implicit differentiation involves a few main steps: differentiation, application of the chain rule, and solving for the desired derivative. Let's explore these steps with examples.

Step 1: Differentiation

Differentiate both sides of the underlying equation with respect to the independent variable. Often, this variable is x. Remember to apply the derivative rules you know, such as the multiplication rule and the power rule.

Step 2: Apply the chain rule

When differentiating terms involving a dependent variable, use the chain rule. For example, if differentiating with respect to x, think of y as a function of x (even if it can't be solved for x explicitly) and apply the chain rule:

d(y²)/dx = 2y * dy/dx

Step 3: Solve for the derivative

After differentiating, you will get an equation involving dy/dx. The goal is to solve this equation for dy/dx, which is the derivative of y with respect to x.

Examples of implicit differentiation

Example 1: Circle equation

Let us first differentiate the presented circle equation:

x² + y² = r²
  1. Differentiating both sides with respect to x gives:
    d(x²)/dx + d(y²)/dx = d(r²)/dx
  2. Uses of derivatives:
    2x + 2y(dy/dx) = 0
  3. Solution of dy/dx:
    2y(dy/dx) = -2x
    dy/dx = -x/y

Example 2: Ellipse equation

Consider the ellipse equation:

x²/a² + y²/b² = 1
  1. Differentiating both sides with respect to x gives:
    d(x²/a²)/dx + d(y²/b²)/dx = d(1)/dx
  2. Uses of derivatives:
    (2x/a²) + (2y/b²)(dy/dx) = 0
  3. Solution of dy/dx:
    (2y/b²)(dy/dx) = -2x/a²
    dy/dx = (-x/a²)/(y/b²) = -bx/ay

Looking at underlying differentiation

Implicit differentiation may be easier to understand by visual examples. Consider a simple circle centered at the origin:

Y X

This represents the implicit function x² + y² = r². If we want to find the slope of the tangent at any point on this circle, we use implicit differentiation. The result of dy/dx = -x/y tells us how the derivatives of x and y on the curve relate to each other.

Implicit differentiation in real-life problems

Physics applications

In physics, implicit differentiation can be useful when dealing with rates of change that cannot be easily separated. For example, consider a scenario where two objects are moving in such a way that their positions are related by an equation. Implicitly differentiating this gives the rate at which the positions change relative to each other.

Economics applications

In economics, implicit functions often arise in constraint optimization problems, such as Lagrange multipliers, where differentiation may be necessary to find equilibria or rates of change within constrained systems.

Exercises and practice problems

To master implicit differentiation, practice with different equations. Below are some exercises:

Exercise 1

Differentiate the following with respect to x:

xy = 7

Solution:

  1. Difference between the two sides:
    x(dy/dx) + y = 0
  2. Solution of dy/dx:
    x(dy/dx) = -y
    dy/dx = -y/x

Exercise 2

Find dy/dx for the following:

sin(xy) = x + y

Solution:

  1. Differentiate both sides using the chain rule:
    cos(xy)(x(dy/dx) + y) = 1 + dy/dx
  2. Rearrange to solve for dy/dx:
    cos(xy)x(dy/dx) + cos(xy)y = 1 + dy/dx
  3. Group the terms containing dy/dx:
    cos(xy)x(dy/dx) - dy/dx = 1 - cos(xy)y
  4. Factor out dy/dx:
    dy/dx(cos(xy)x - 1) = 1 - cos(xy)y
  5. Solve for dy/dx:
    dy/dx = (1 - cos(xy)y)/(cos(xy)x - 1)

Conclusion

Implicit differentiation is an important concept in calculus that enhances our ability to deal with complex relationships between variables. By following the steps of differentiation, correctly applying the chain rule, and solving for the desired derivative, we can tackle a wide range of mathematical and real-life problems. Practicing and applying these principles will build proficiency and intuition in dealing with implicit functions.


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