91Ó°ÊÓ

When dealing with functions, the second derivative tells us about the concavity of the graph of a function and the rate of change of the first derivative. It provides insight into how the slope of the curve is changing.
For the given equation, we started by differentiating the equation of the circle, \(x^2 + y^2 = 36\), implicitly to find the first derivative, \(dy/dx\). The first derivative was determined as \(-x/y\).
To find the second derivative, we differentiated \(dy/dx = -x/y\) once more, using the quotient rule, ending up with \(d^2y/dx^2 = -2x/y^2\).
The second derivative here \(d^{2} y / d x^{2}\) helps show how the slope \(dy/dx\) changes, which can reveal more about the curve's bending nature.

The quotient rule is a valuable tool when differentiating a function that is a ratio of two functions. If you have a function \(\frac{u}{v}\), where both \(u\) and \(v\) are functions of \(x\), the quotient rule states that the derivative is:
\[ \left(\frac{u}{v}\right)' = \frac{v \cdot u' - u \cdot v'}{v^2} \]
In the exercise, applying the quotient rule was essential when differentiating \(dy/dx = -x/y\) to find the second derivative. This step involves setting \(u = -x\) and \(v = y\), leading to:

• Derive \(u\), which gives \(u' = -1\).
• Derive \(v\), which gives \(v' = dy/dx = -x/y\).

Substituting these derivatives into the quotient rule formula resulted in the second derivative \(-2x/y^2\). This rule is crucial when dealing with ratios of functions in calculus, simplifying complex expressions into manageable forms.

Implicit differentiation is key when a relationship between \(x\) and \(y\) isn't explicitly given, like in the equation of a circle \(x^2 + y^2 = 36\). Here, \(y\) is defined implicitly rather than as an explicit function of \(x\).
To perform implicit differentiation, apply the derivative to each term while treating \(y\) as an implicit function of \(x\). This means we differentiate \(y\) with respect to \(x\) every time we see it, often resulting in extra terms.
In our task, differentiating both sides of \(x^2 + y^2=36\) yielded \(2x + 2y(dy/dx) = 0\). Solving for \(dy/dx\) gave \(-x/y\), our first derivative. This approach allows us to find derivatives of functions not easily expressed solely in terms of one variable, giving insights that wouldn't be easily attainable otherwise.