Problem 90
Question
Find a formula for \(\frac{d}{d x}[f(x)]^{-1}\) by writing it as \(\frac{d}{d x}\left[\frac{1}{f(x)}\right]\) and using the Quotient Rule. Be sure to simplify your answer.
Step-by-Step Solution
Verified Answer
\( \frac{d}{dx}[f(x)]^{-1} = \frac{-f'(x)}{[f(x)]^2} \).
1Step 1: Express the Function as a Quotient
To begin, note that \( [f(x)]^{-1} \) is equivalent to \( \frac{1}{f(x)} \). This allows us to prepare the expression for differentiation using the quotient rule.
2Step 2: Recall the Quotient Rule
The quotient rule states that if you have a function expressed as \( \frac{u(x)}{v(x)} \), then its derivative is \( \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2} \). For our function \( \frac{1}{f(x)} \), we identify \( u(x) = 1 \) and \( v(x) = f(x) \).
3Step 3: Differentiate Using the Quotient Rule
Apply the quotient rule using \( u(x) = 1 \) and \( v(x) = f(x) \). This gives\[ \frac{d}{dx} \left[ \frac{1}{f(x)} \right] = \frac{f(x) \cdot \frac{d}{dx}[1] - 1 \cdot \frac{d}{dx}[f(x)]}{[f(x)]^2} \].Since the derivative of a constant \(1\) is \(0\), the formula simplifies to:\[ \frac{f(x) \cdot 0 - 1 \cdot f'(x)}{[f(x)]^2} = \frac{-f'(x)}{[f(x)]^2} \].
4Step 4: Simplify the Expression
The expression simplifies to:\[ \frac{-f'(x)}{[f(x)]^2} \].This is the derivative of \( [f(x)]^{-1} \) with respect to \( x \).
Key Concepts
Quotient RuleMathematical FunctionsDifferentiation Techniques
Quotient Rule
When dealing with derivatives of quotient functions, the Quotient Rule is an essential tool. Understanding how to apply it effectively can simplify complex expressions.
The Quotient Rule provides a method for finding the derivative of a division of two differentiable functions.
If a function is expressed as \( \frac{u(x)}{v(x)} \), where both functions \( u(x) \) and \( v(x) \) are differentiable, then the derivative of this quotient is given by:
\[ \frac{d}{dx} \left[ \frac{u(x)}{v(x)} \right] = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2}. \]
In simple terms, you:
The Quotient Rule provides a method for finding the derivative of a division of two differentiable functions.
If a function is expressed as \( \frac{u(x)}{v(x)} \), where both functions \( u(x) \) and \( v(x) \) are differentiable, then the derivative of this quotient is given by:
\[ \frac{d}{dx} \left[ \frac{u(x)}{v(x)} \right] = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2}. \]
In simple terms, you:
- Multiply the derivative of the numerator \( u'(x) \) by the denominator \( v(x) \).
- Subtract the product of the numerator \( u(x) \) and the derivative of the denominator \( v'(x) \).
- Divide the entire result by \( v(x)^2 \).
Mathematical Functions
Mathematical functions are expressions that relate one quantity to another.
When working on derivatives in calculus, understanding the nature of functions is crucial.
A function maps inputs (often called \( x \) values) to respective outputs through a specific rule. For instance, in \( f(x) = x^2 \), if \( x \) is 3, then \( f(x) \) results in 9.
There are several types of functions, including:
When working on derivatives in calculus, understanding the nature of functions is crucial.
A function maps inputs (often called \( x \) values) to respective outputs through a specific rule. For instance, in \( f(x) = x^2 \), if \( x \) is 3, then \( f(x) \) results in 9.
There are several types of functions, including:
- Linear functions: These have constant rates of change, like \( f(x) = 3x + 2 \).
- Polynomial functions: Include terms like \( x^2 \), \( x^3 \), etc., such as \( f(x) = x^3 - 4x + 5 \).
- Trigonometric functions: Functions like sine and cosine that deal with angles.
- Exponential functions: Involving constant bases raised to variable powers, such as \( f(x) = 2^x \).
- Rational functions: Ratios of polynomials, like \( f(x) = \frac{x^2+1}{x-3} \).
Differentiation Techniques
Differentiation is a key concept of calculus focused on finding the rate of change of functions.
It’s used to determine how a function's output changes with respect to its input.
There are several techniques to perform differentiation:
It’s used to determine how a function's output changes with respect to its input.
There are several techniques to perform differentiation:
- Power Rule: Useful for any function of the form \( f(x) = x^n \). The derivative is \( nx^{n-1} \).
- Product Rule: If you have two multiplied functions \( u(x) \) and \( v(x) \), the derivative is \( u'(x)v(x) + u(x)v'(x) \).
- Chain Rule: Used for composite functions. If \( y = g(f(x)) \), then the derivative is \( g'(f(x)) \cdot f'(x) \).
- Quotient Rule: As previously mentioned, this is for divisions of functions.
Other exercises in this chapter
Problem 89
Find a formula for \(\frac{d}{d x}[f(x)]^{2}\) by writing it as \(\frac{d}{d x}[f(x) f(x)]\) and using the Product Rule. Be sure to simplify your answer.
View solution Problem 90
True or False: If \(\lim _{x \rightarrow 2} f(x)=7\), then \(\lim _{x \rightarrow 2^{+}} f(x)=7\)
View solution Problem 91
True or False: \(\lim _{x \rightarrow 1} \frac{x^{2}-1}{x-1}=\frac{\lim _{x \rightarrow 1}\left(x^{2}-1\right)}{\lim (x-1)}\)
View solution Problem 91
Beverton-Holt Recruitment Curve Some organisms exhibit a density-dependent mortality from one generation to the next. Let \(R>1\) be the net reproductive rate (
View solution