Problem 2 - A guide to MATLAB: For Beginners and Experienced Users

Question

Find the derivatives of the following functions, and if possible, simplify each answer: (a) \(f(x)=6 x^{3}-5 x^{2}+2 x-3\) (b) \(f(x)=\frac{2 x-1}{x^{2}+1}\) (c) \(f(x)=\sin \left(3 x^{2}+2\right)\) (d) \(f(x)=\arcsin (2 x+3)\) (e) \(f(x)=\sqrt{1+x^{4}}\) (f) \(f(x)=x^{r}\)

Step-by-Step Solution

Answer

(a) \(f'(x) = 18x^2 - 10x + 2\), (b) \(f'(x) = \frac{-2x^2 + 2x + 2}{(x^2+1)^2}\), (c) \(6x \cos(3x^2+2)\), (d) \(\frac{2}{\sqrt{1-(2x+3)^2}}\), (e) \(\frac{2x^3}{\sqrt{1+x^4}}\), (f) \(rx^{r-1}\).

Step 1: Differentiate a Polynomial Function (Part a)

To find the derivative of the polynomial function \(f(x) = 6x^3 - 5x^2 + 2x - 3\), apply the power rule \(\frac{d}{dx}(x^n) = nx^{n-1}\). Differentiating each term separately, we have:- \(\frac{d}{dx}(6x^3) = 18x^2\)- \(\frac{d}{dx}(-5x^2) = -10x\)- \(\frac{d}{dx}(2x) = 2\)- \(\frac{d}{dx}(-3) = 0\)Therefore, the derivative \(f'(x) = 18x^2 - 10x + 2\).

Step 2: Differentiate a Rational Function (Part b)

For the function \(f(x)=\frac{2x-1}{x^2+1}\), use the quotient rule \(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\cdot u' - u\cdot v'}{v^2}\), where \(u = 2x - 1\) and \(v = x^2 + 1\).dash Derivatives: \(u' = 2\), \(v' = 2x\).Plug these into the quotient rule:\[f'(x) = \frac{(x^2+1)(2) - (2x-1)(2x)}{(x^2+1)^2}\]Simplify the expression:\[f'(x) = \frac{2x^2 + 2 - (4x^2 - 2x)}{(x^2+1)^2} = \frac{-2x^2 + 2x + 2}{(x^2+1)^2}\]

Step 3: Differentiate a Composite Function using the Chain Rule (Part c)

To differentiate \(f(x)=\sin(3x^2+2)\), apply the chain rule: \(\frac{d}{dx}(\sin u) = \cos u \cdot \frac{du}{dx}\). Here, \(u = 3x^2 + 2\), thus \(\frac{du}{dx} = 6x\).The derivative is:\[f'(x) = \cos(3x^2+2) \cdot 6x = 6x \cos(3x^2+2)\]

Step 4: Differentiate an Inverse Trigonometric Function (Part d)

For \(f(x)=\arcsin(2x+3)\), the derivative of \(\arcsin(u)\) is \(\frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}\). Here, \(u = 2x + 3\), \(\frac{du}{dx} = 2\).The derivative is:\[f'(x) = \frac{2}{\sqrt{1 - (2x+3)^2}}\]

Step 5: Differentiate a Function involving Square Roots (Part e)

For \(f(x)=\sqrt{1+x^4}\), rewrite it as \((1 + x^4)^{1/2}\) and apply the chain rule:\[f'(x) = \frac{1}{2}(1+x^4)^{-1/2} \cdot \frac{d}{dx}(1 + x^4)\]Derivatives: \(\frac{d}{dx}(1 + x^4) = 4x^3\).The derivative is:\[f'(x) = \frac{4x^3}{2\sqrt{1+x^4}} = \frac{2x^3}{\sqrt{1+x^4}}\]

Step 6: Differentiate an Exponential Function with Power (Part f)

For \(f(x)=x^r\), use the power rule: \(\frac{d}{dx}(x^r) = rx^{r-1}\).The derivative is:\[f'(x) = rx^{r-1}\]

Key Concepts

Derivative

The concept of a derivative is foundational in calculus. It represents how a function changes as its input changes. Simply put, a derivative gives the rate at which one quantity changes with respect to another. In mathematical terms, the derivative of a function \( f(x) \) at a point \( x \) is given by: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] This formula is known as the limit definition of a derivative. Computing a derivative can tell us things like the slope of a line tangent to the function at a point, which can help in understanding how the function behaves locally.

• The derivative can indicate speed and velocity in physical problems.
• It also helps in optimizing areas, such as maximizing profit in economic problems.

Understanding derivatives is crucial for solving problems involving rates of change and making predictions.

Quotient Rule

The quotient rule is a method used to differentiate functions that are fractions, specifically when one function is divided by another. If you have a function expressed as \( \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), then the quotient rule is stated as follows:\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2} \]This rule allows us to find the derivative of complex rational functions without having to simplify them first.To use the quotient rule:

• Identify the "top" function \( u \) and "bottom" function \( v \).
• Differentiate \( u \) to get \( u' \) and differentiate \( v \) to get \( v' \).
• Substitute \( u \), \( u' \), \( v \), and \( v' \) into the quotient rule formula.

This method simplifies differentiation of functions like \( \frac{2x-1}{x^2+1} \), improving accuracy in computational tasks.

Chain Rule

The chain rule is a technique used in calculus to differentiate composite functions. A composite function is where one function is nested inside another. For instance, in the function \( \sin(3x^2+2) \), the outer function is the sine, and the inner function is \( 3x^2+2 \).The chain rule can be applied using the following formula:\[ \frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x) \]Here’s how to apply it:

• Identify the outer function \( f \) and the inner function \( g \).
• Differentiating the outer function \( f \), leaving the inner function \( g(x) \) unchanged.
• Multiply by the derivative of the inner function \( g'(x) \).

This method is necessary for tackling more complex derivatives, enabling calculations like finding \( f'(x) = 6x \cos(3x^2+2) \), a crucial step in advanced calculus problems.

Polynomial Derivative

Calculating the derivative of a polynomial function is typically straightforward due to the power rule. For a polynomial function of the form \( f(x) = ax^n \), the derivative is found using:\[ \frac{d}{dx}(ax^n) = n \cdot ax^{n-1} \]This allows for quick computation of derivatives of terms individually:

• For \( 6x^3 \), the derivative is \( 18x^2 \).
• For \( -5x^2 \), the derivative is \( -10x \).
• For linear terms like \( 2x \), the derivative simply is \( 2 \).

Constants like \( -3 \) become zero as they don’t change with \( x \). By applying the power rule term by term, we easily find the overall derivative of polynomials, vital in solving equations involving rates and ensuring successful problem-solving in calculus.