Question

If \(\mathrm{y}^{2}=\mathrm{p}(\mathrm{x})\) is polynomial function more than 3 degree then \(2(\mathrm{~d} / \mathrm{dx})\left|\mathrm{y}^{3} \mathrm{y}_{2}\right|=\) (a) \(\mathrm{p}^{\prime}(\mathrm{x}) \cdot \mathrm{p}^{\prime \prime}(\mathrm{x})\) (b) \(\mathrm{p}(\mathrm{x}) \cdot \mathrm{p}^{\prime \prime}(\mathrm{x})\) (c) \(\mathrm{p}^{\prime}(\mathrm{x}) \cdot \mathrm{p}^{\prime \prime \prime}(\mathrm{x})\) (d) \(p(x) \cdot p^{\prime \prime \prime}(x)\)

Short answer

The given problem cannot be solved with the information provided in the exercise. The presence of the absolute value function makes it impossible to match the expression with the given options.

Step-by-step solution

Finding the first derivative of y

First, we will implicitly differentiate the given equation with respect to \(x\): \(y^2 = p(x)\) Differentiating both sides with respect to x, we get: \(2y\frac{dy}{dx} = p'(x)\) Now, solving for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{p'(x)}{2y}\)

Finding the second derivative of y

Next, let's differentiate the first derivative with respect to \(x\) to find the second derivative \(\frac{d^2y}{dx^2}\): \(\frac{d}{dx}\left(\frac{p'(x)}{2y}\right) = \frac{p''(x) \cdot 2y - p'(x) \cdot 2y\frac{dy}{dx}}{(2y)^2}\) But we know that \(\frac{dy}{dx} = \frac{p'(x)}{2y}\). So, substitute it into the equation: \(\frac{d^2y}{dx^2} = \frac{p''(x) \cdot 2y - p'(x) \cdot p'(x)}{(2y)^2}\).

Finding the product of y^3 and the second derivative

Now let's find the product of \(y^3\) and its second derivative, and take the absolute value: \( |y^3\frac{d^2y}{dx^2}| = \left| \frac{y^3(p''(x) \cdot 2y - p'(x) \cdot p'(x))}{(2y)^2} \right|\)

Multiply the result by 2 and differentiate it with respect to x

Next, we will multiply the result from Step 3 by 2 and differentiate with respect to \(x\): \(\frac{d}{dx}\left( 2|y^3\frac{d^2y}{dx^2}| \right)= \frac{d}{dx}\left( |2y^3(p''(x) \cdot 2y - p'(x) \cdot p'(x))| \right)\) However, due to the presence of the absolute value, we can't provide a specific result without having more information about the polynomial function or its derivatives. Therefore, it is not possible to match the expression with any of the given options. In conclusion, the given problem cannot be solved with the information provided in the exercise. The presence of the absolute value function makes it impossible to match the expression with the given options.

Key concepts

Polynomial Function

A polynomial function is a mathematical expression that consists of terms made up of variables raised to whole number exponents and multiplied by coefficients. These functions are known for their smooth and continuous curves and have different degrees depending on the highest power of the variable. For example:

• A linear polynomial, such as \(ax + b\), is of degree 1.
• A quadratic polynomial, like \(ax^2 + bx + c\), is of degree 2.
• A cubic polynomial, \(ax^3 + bx^2 + cx + d\), is of degree 3.

In the original exercise, you encounter a polynomial function \(p(x)\) with a degree higher than 3 embedded within the equation \(y^2 = p(x)\). Understanding this helps set the stage for finding derivatives, as the complexity and behavior of the polynomial will impact the differentiation process.

Higher-Order Derivatives

The process of finding the rate at which a function changes is known as differentiation. While the first derivative gives the slope or the rate of change of the function, higher-order derivatives provide additional insights into how these changes themselves change.

• The second derivative \(\frac{d^2y}{dx^2}\) assesses the concavity of the function and can indicate inflection points where the function's direction of curvature changes.
• Higher-order derivatives can become increasingly complex, especially when dealing with functions like polynomial functions with degrees higher than three.

Applying this concept in the exercise, after determining the first derivative \(\frac{dy}{dx}\), you find the second derivative \(\frac{d^2y}{dx^2}\). This involves differentiating the first derivative again, which requires applying product and chain rules due to its composition of functions.

Differentiation Rules

Differentiation rules are essential tools for finding derivatives of functions efficiently. The rules simplify the process of deriving expressions, especially when dealing with polynomial functions or functions composed of multiple terms.

• The power rule states: If \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\).
• The product rule is helpful when differentiating products of two functions: If \(u(x)\) and \(v(x)\) are functions, then \((uv)' = u'v + uv'\).
• The chain rule is useful for differentiating composite functions: If a function \(y = g(f(x))\), then \(\frac{dy}{dx} = g'(f(x)) \, f'(x)\).

In the original exercise, these differentiation rules are key to finding derivatives implicitly. Implicit differentiation is a technique used here since \(y\) is not isolated; the equation is differentiated with respect to \(x\) while treating \(y\) as a function of \(x\). Understanding these rules allows us to navigate the complexity presented by the polynomial \(p(x)\) and solve for derivatives in the problem efficiently.