91Ó°ÊÓ

In calculus, the chain rule is a fundamental tool for differentiating composite functions. When a function is nested within another function, the chain rule allows us to differentiate efficiently. Imagine you have a situation where one variable depends on another, which in turn depends on a third variable. This is where the chain rule shines.
For example, if we have a function of the form \( y = g(f(x)) \), the derivative \( dy/dx \) can be found by multiplying the derivative of the outer function \( g \) with respect to \( f \) by the derivative of the inner function \( f \) with respect to \( x \).

• The formula: \( \frac{dy}{dx} = \frac{dg}{df} \cdot \frac{df}{dx} \).
• This rule helps break down complex problems into manageable steps.

In implicit differentiation, the chain rule is vital because you often hold one variable constant while differentiating with respect to another.

The quotient rule is a method for finding the derivative of a fraction, that is, a ratio of two functions. When you have a function \( f(x) = \frac{u(x)}{v(x)} \), the derivative, \( f'(x) \), is found using:

• \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \)

This provides a structured way to calculate derivatives when the function is a quotient.

Steps to Use the Quotient Rule:

• Differentiating the numerator \( u \) and denominator \( v \) separately.
• Substitute into the formula.
• Ensure that you square the denominator.

This rule is essential in tasks like implicit differentiation, where a function may appear as a fraction of other expressions.

An undefined derivative occurs when the derivative of a function doesn't exist at a certain point. This can happen for several reasons:

• The function isn't continuous.
• There's a sharp corner or cusp in the graph.
• A division by zero arises in the derivative expression.

In the provided exercise, the derivative is undefined at the point \((7,0)\) because substituting these values into the expression results in a division by zero.
This typically indicates a vertical tangent line or a different irregularity at that specific point, thus making it so that \( dy/dx \) cannot be determined there.

Evaluation at a point in derivatives involves substituting specific \( x \) and \( y \) values into the derivative expression to find the slope of the tangent line at that point. This is often done after applying implicit or explicit differentiation.
To evaluate:

• First, differentiate the equation.
• Next, substitute the given point's \( x \) and \( y \) values into the derivative.

In our case, evaluating the derivative at \((7,0)\) led to a division by zero, indicating an undefined slope.
Understanding this helps identify where functions may have vertical tangents or be otherwise not differentiable, giving valuable insights into the graph's behavior at that point.