91Ó°ÊÓ

When dealing with curves defined by equations that aren't solely expressed as functions of one variable, implicit differentiation is a useful technique. It helps find the derivative when it's challenging to express a variable explicitly. For example, in the equation \( \sin y + 5x = y^2 \), neither \( x \) nor \( y \) is isolated.

By differentiating each term with respect to \( x \), we can solve for \( \frac{dy}{dx} \) even though \( y \) is within a sine function. The process involves treating \( y \) as a function of \( x \) and applying the chain rule.

Core steps of implicit differentiation include:

• Differentiate all terms with respect to \( x \).
• Watch for terms involving \( y \), apply the chain rule and attach \( \frac{dy}{dx} \) to them.
• Solve for \( \frac{dy}{dx} \) once the differentiation is complete.

This technique is essential for curves where variables are interconnected in complex ways.

Finding the equation of a tangent line involves calculating the slope at a specific point on the curve and then using that to form the equation. The point-slope form of the equation of a line is particularly useful here.

To find the slope of the tangent line at a given point, one must:

• Use implicit differentiation to find \( \frac{dy}{dx} \).
• Substitute the coordinates of the given point into \( \frac{dy}{dx} \) to find the slope.

In our example, the slope \( \frac{5}{2\pi + 1} \) is calculated by plugging in \( y = \pi \).

With the point \( \left(\frac{\pi^2}{5}, \pi\right) \) and the slope, the equation becomes: \[y - \pi = \frac{5}{2\pi + 1}(x - \frac{\pi^2}{5})\]This equation represents the tangent line at the specified point.

Verifying a point on a curve is a straightforward but crucial step in understanding the relationship between the curve and the given point. To verify, one substitutes the point's coordinates into the given curve equation.

For the equation \( \sin y + 5x = y^2 \), checking if the point \( \left( \frac{\pi^2}{5}, \pi \right) \) lies on it involves:

• Substituting \( x = \frac{\pi^2}{5} \) and \( y = \pi \) into the equation.
• Ensuring both sides of the equation match.

In this scenario, substituting gives \( \sin(\pi) + 5 \times \frac{\pi^2}{5} = \pi^2 \), which confirms that the point is on the curve as both sides equal \( \pi^2 \).

This process ensures the integrity of subsequent analysis like finding tangent lines.

The chain rule is a powerful calculus tool used when differentiating compositions of functions. It particularly comes into play during implicit differentiation and is necessary whenever a function inside another is differentiated.

In the context of \( \sin y + 5x = y^2 \), the chain rule is applied:

• When differentiating \( \sin y \), as \( y \) is a function of \( x \).
• Also, \( y^2 \) when \( y \) itself is differentiable, necessitating \( \frac{dy}{dx} \).

This results in expressions like \( \cos y \cdot \frac{dy}{dx} \) ensuring that every derivative respects the dependency of \( y \) on \( x \).

Mastering the chain rule is vital for any problem involving multiple composed functions, such as those encountered in implicit differentiation.