Chapter 2: Problem 2
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) \(f(x)=x^{r}\)
Short Answer
Expert verified
(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 by step solution
01
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 \(f'(x) = 18x^2 - 10x + 2\).
02
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:\[f'(x) = \frac{2x^2 + 2 - (4x^2 - 2x)}{(x^2+1)^2} = \frac{-2x^2 + 2x + 2}{(x^2+1)^2}\]
03
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:\[f'(x) = \cos(3x^2+2) \cdot 6x = 6x \cos(3x^2+2)\]
04
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:\[f'(x) = \frac{2}{\sqrt{1 - (2x+3)^2}}\]
05
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:\[f'(x) = \frac{4x^3}{2\sqrt{1+x^4}} = \frac{2x^3}{\sqrt{1+x^4}}\]
06
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:\[f'(x) = rx^{r-1}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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 \( g'(x) \).
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 \( 2 \).