91Ó°ÊÓ

The chain rule is a powerful tool in calculus that helps us differentiate composite functions. When you have a function within another function, the chain rule allows you to find the derivative efficiently. For example, if we have a function \( y = f(u) \) and \( u = g(x) \), then the derivative \( \frac{dy}{dx} \) can be found using the chain rule as follows:

• First, differentiate the outer function with respect to the inner function: \( \frac{df}{du} \).
• Then, multiply by the derivative of the inner function with respect to \( x \): \( \frac{dg}{dx} \).

So, \( \frac{dy}{dx} = \frac{df}{du} \cdot \frac{dg}{dx} \). In our exercise, we applied the chain rule when differentiating \( \sin y \) with respect to \( x \). Since \( y \) is a function of \( x \), we use the chain rule to get \( \cos y \cdot \frac{dy}{dx} \). This takes into account both the derivative of \( \sin y \) and the fact that \( y \) itself changes with \( x \).

To differentiate a function that is the quotient of two other functions, we use the quotient rule. This rule is crucial when dealing with divisions of functions in calculus. If you have a function that looks like \( \frac{u}{v} \), where both \( u(x) \) and \( v(x) \) are functions of \( x \), then the derivative is given by:

• The formula for the quotient rule is: \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \).

In our problem, to find the second derivative \( \frac{d^2y}{dx^2} \), we applied the quotient rule. We had to differentiate the expression \( \frac{-3x^2}{\cos y - 1} \) which is a ratio of two functions. We calculated:

• The derivative of the numerator \(-3x^2\), yielding \(-6x\).
• The derivative of the denominator \( \cos y - 1 \), taking \( \cos y \)'s derivative as \(-\sin y\) times \( \frac{dy}{dx} \).

Then, by substituting these derivatives into the quotient rule formula, we were able to solve for \( \frac{d^2y}{dx^2} \).

The second derivative measures how the rate of change (the first derivative) is itself changing. It provides insights into the concavity and acceleration of a function, helping us understand the function's curvature. To compute the second derivative \( \frac{d^2y}{dx^2} \), we first found the first derivative using implicit differentiation. Then, we differentiated again.

• In doing so, it's essential to apply the rules of differentiation consistently, involving steps such as the quotient rule or chain rule, as needed.
• The expression \( \frac{-3x^2}{\cos y - 1} \) was differentiated using the quotient rule, as it is a division of two functions.
• Substituting the result of the first derivative \( \frac{dy}{dx} = \frac{-3x^2}{\cos y - 1} \) into our calculation helped simplify the process.

The final form of the second derivative, \( \frac{d^2y}{dx^2} = \frac{-6x(\cos y - 1) + 9x^4 \sin y}{(\cos y - 1)^3} \), gives us the required expression that highlights how changes in \( x \) affect the curvature of our graph.