91Ó°ÊÓ

In calculus, the second derivative is the derivative of the derivative of a function. It provides information about the curvature of the function's graph and how it changes. It can tell us whether the graph of the function is concave up or concave down at a certain point and is often used to find acceleration in physics, as acceleration is the second derivative of position with respect to time.

In our exercise, finding the second derivative \(\frac{d^{2} y}{d x^{2}}\) requires differentiation of the first derivative. Since this involves an implicit function (where y is not isolated on one side), we first found \(\frac{dy}{dx}\) by implicitly differentiating the given equation, and then we differentiated \(\frac{dy}{dx}\) again to find \(\frac{d^{2} y}{d x^{2}}\).

The chain rule is a fundamental differentiation rule in calculus used when taking the derivative of a composite function. The rule can be viewed as taking the derivative of the outside function and multiplying it by the derivative of the inside function. In symbolic form, if \(f(x) = g(h(x))\), then \(f'(x) = g'(h(x)) \cdot h'(x)\).

In our exercise, when finding the first derivative, the term \(x^{2} y\) requires the use of the chain rule, because \(y\) is a function of \(x\), and \(y\) itself is being multiplied by \(x^{2}\). This is why \(x^{2} \frac{dy}{dx}\) appears in the differentiated equation.

The quotient rule is another differentiation technique that is used when you need to find the derivative of a fraction. If you have a function \(u(x)\) divided by another function \(v(x)\), the quotient rule states the derivative is \(\frac{u'(x) v(x) - u(x)v'(x)}{[v(x)]^{2}}\).

Applying this to our problem involves finding the second derivative \(\frac{d^{2} y}{d x^{2}}\): when you differentiate the first derivative \(\frac{dy}{dx}\) which was in a quotient form, you use the quotient rule. Each term in the numerator is processed by considering the differentiation of \(\cos x + 2xy\) as well as the square of the denominator \(x^{2}\).

When functions are defined explicitly, y is isolated on one side and x on the other, such as \(y = f(x)\). However, an implicit function is where y cannot be easily isolated, and both x and y are mixed within the equation. It’s still possible to find the derivative of y with respect to x by using implicit differentiation, where we treat y as a function of x throughout differentiation.

In our initial exercise \(\sin x + x^{2} y = 10\), y is not isolated. To differentiate with respect to x, we consider y as an implicit function. This means each time we encounter y during differentiation, we treat it as a function of x that's being differentiated, resulting in additional terms involving \(\frac{dy}{dx}\).