Question

Differentiate the following funetions: \(y=\frac{1}{\sqrt{3-2 x+4 x^{2}}}\)

Short answer

Answer: The derivative of the given function y with respect to x is: \(\frac{dy}{dx} = \frac{(2-8x)}{2\sqrt{(3-2x+4x^{2})^{3}}}\)

Step-by-step solution

Identify the inner and outer functions

The inner function is the quadratic function inside the square root, which we can denote as u(x) = 3-2x+4x^2. The outer function is the reciprocal of the square root, which we can denote as y(v) = \(\frac{1}{\sqrt{v}}\), where v is a function of x.

Differentiate the inner function with respect to x

Differentiate u(x) = 3-2x+4x^2 with respect to x to find its derivative, du/dx. \(\frac{du}{dx} = -2 + 8x\)

Differentiate the outer function with respect to v

Differentiate y(v) = \(\frac{1}{\sqrt{v}}\) with respect to v to find its derivative, dy/dv. \(\frac{dy}{dv} = \frac{-1}{2v^{\frac{3}{2}}}\)

Apply the chain rule

According to the chain rule, the derivative of y with respect to x is given by dy/dx = dy/dv * du/dx. Substitute the expressions for du/dx and dy/dv in the chain rule, and remember that v is a function of x. \(\frac{dy}{dx} = \frac{-1}{2(3-2x+4x^{2})^{\frac{3}{2}}} \cdot (-2+8x)\)

Simplify the expression

Combine the terms and simplify the expression for dy/dx. \(\frac{dy}{dx} = \frac{-(-2+8x)}{2\sqrt{(3-2x+4x^{2})^{3}}}\) So, the derivative of the given function y with respect to x is: \(\frac{dy}{dx} = \frac{(2-8x)}{2\sqrt{(3-2x+4x^{2})^{3}}}\)

Key concepts

Differentiation

Differentiation is a core concept in calculus that focuses on finding the rate at which a function changes. When we differentiate a function, we are essentially looking at how the dependent variable y changes with respect to the independent variable x. The result of differentiation is called the derivative.

The process involves taking a given function and computing its derivative, which gives us a new function that describes the instantaneous rate of change. This is especially useful for understanding how functions behave and how their graphs evolve.

• Differentiation provides insights into slopes of tangent lines at any point on a curve.
• It helps in identifying areas of increase and decrease of functions.
• It's crucial for solving problems in physics and engineering where change over time or space is analyzed.

In the context of our problem, differentiation allows us to calculate how the function \(y=\frac{1}{\sqrt{3-2x+4x^{2}}}\) changes as x varies.

Chain Rule

The chain rule is a powerful method in calculus for differentiating composite functions, those where one function is nested inside another. Calculus students often encounter situations where the chain rule is the most efficient approach to finding a derivative.

Using the chain rule involves these steps:

• Identify the inner and outer functions. The inner function here is \(u(x) = 3-2x+4x^{2}\), while the outer function is \(y(v) = \frac{1}{\sqrt{v}}\).
• Differentiate the inner function, \(\frac{du}{dx}\), and the outer function with respect to the inner variable, \(\frac{dy}{dv}\).
• Finally, combine these derivatives to obtain the derivative of the composite function with respect to x using the formula: \(\frac{dy}{dx} = \frac{dy}{dv} \cdot \frac{du}{dx}\).

This rule is invaluable when dealing with more complex functions as it breaks down the differentiation process into more manageable parts.

Functions

Functions are foundational elements in mathematics and calculus, as they represent relationships between two quantities. A function maps an input (usually denoted as x) to a unique output (denoted as y). The way functions behave provides insights into countless mathematical problems and real-world scenarios.

Functions are useful for modeling a wide array of phenomena:

• Physics uses functions to describe motions and forces.
• Economics and finance rely on functions for modeling trends and predictions.
• Biology uses them to track growth and decay in ecosystems.

In the given exercise, we are working with a function \(y=\frac{1}{\sqrt{3-2x+4x^{2}}}\), which is a composite of a square root function and a quadratic function. Understanding how to differentiate this function involves appreciating both the components and their interaction through differentiation techniques.

Quadratic Functions

Quadratic functions are polynomial functions of degree two, and they take the form \(ax^{2} + bx + c\), where a, b, and c are constants. These functions are significant in mathematics because they model parabolic behavior, which is encountered in numerous scientific and engineering problems.

Quadratic functions have distinctive properties:

• They graph as parabolas, which can open upwards or downwards depending on the sign of the \(a\) coefficient.
• The vertex form or the standard form gives insights into the vertex of the parabola, which is its highest or lowest point.
• They have axis of symmetry, typically a vertical line that passes through the vertex.

In our exercise, the inner function \(u(x) = 3-2x+4x^{2}\) is quadratic. Differentiating this function gives us a linear derivative \(-2 + 8x\), which is part of computing the derivative of the composite function using the chain rule.