Narrowing Down to One Variable with the Help of Implicit Differentiation

How to solve them?

Implicit differentiation is a valuable calculus tool, especially useful when dealing with equations that involve both \( x \) and \( y \) in a way that makes it difficult to express \( y \) explicitly as a function of \( x \) . Here’s how you can approach it:

• Start with the Equation: Begin with your given equation where \( x \) and \( y \) are mixed. For example, an equation like \( x^2 + y^2 = 1 \). Differentiate Both Sides: Treat the equation like any other function you would differentiate, remembering that you are differentiating with respect to \( x \) . This means every time you differentiate a term with \( y \) in it, you have to multiply by \( \frac{dy}{dx} \) because of the chain rule.
• Apply Derivative Rules: Use all the derivative rules you know (like the power rule, product rule, etc.) as you work through each term of the equation. For instance, when differentiating \( x^2 \) , it’s straightforward. But for \( y^2 \) , you differentiate as usual and then multiply by \( \frac{dy}{dx} \).
• Rearrange and Solve for \( \frac{dy}{dx} \) : Once you’ve differentiated all terms, rearrange the equation to solve for \( \frac{dy}{dx} \) . This might involve collecting all the terms with \( \frac{dy}{dx} \)
  • on one side of the equation and then isolating \( \frac{dy}{dx} \) .
• Interpret Your Result: The \( \frac{dy}{dx} \) you’ve solved is the derivative of \( y \) with respect to \( x \) . It tells you how \( y \) changes as \( x \) changes within the context of your original equation. Implicit differentiation is particularly handy in cases where solving for \( y \) explicitly would be too cumbersome or impossible. It’s a skill that’s especially useful in fields like physics, where many relationships between variables are not straightforward. This method ensures that students can find the derivatives they need for further analysis in calculus problems, despite the complexity of the equations involved.

Example:

Find the derivative \( \frac{dy}{dx} \) of the circle defined by the equation \( x^2 + y^2 = 25 \) using implicit differentiation.

Solution:

The equation of the circle is \( x^2 + y^2 = 25 \). Differentiating each term with respect to \( x\) gives: \( 2x + 2y \frac{dy}{dx} = 0 \)

Here, \( 2x \) is the derivative of \( x^2 \) and \( 2y \frac{dy}{dx} \) is the derivative of \( y^2 \), applying the chain rule.

Rearrange the equation to isolate \( \frac{dy}{dx} \): \( 2y \frac{dy}{dx} = -2x \) and then \( \frac{dy}{dx} = \frac{-2x}{2y} \). Simplify to get \( \frac{dy}{dx} = \frac{-x}{y} \).

So the derivative \( \frac{dy}{dx} \) represents the slope of the tangent to the circle at any point \( (x,y) \) on the circle.