Problem 12
Question
In Problems 1-28, differentiate the functions with respect to the independent variable. $$ f(x)=\frac{\left(1-3 x^{2}\right)^{3}}{\left(3-x^{2}\right)^{2}} $$
Step-by-Step Solution
Verified Answer
Differentiate using the quotient rule and chain rule; the derivative is complex.
1Step 1: Identify the Differentiation Rule Needed
The function is given as a fraction of two functions, suggesting the quotient rule will be necessary. Each component of the fraction involves powers, indicating that the chain rule might also be applied.
2Step 2: Apply the Quotient Rule
The quotient rule for derivatives is \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \), where \( u = (1-3x^2)^3 \) and \( v = (3-x^2)^2 \). Calculate \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) next.
3Step 3: Differentiate the Numerator using the Chain Rule
Let \( u = (1-3x^2)^3 \). Apply the chain rule: \( \frac{du}{dx} = 3(1-3x^2)^2 \cdot \frac{d}{dx}(1-3x^2) = 3(1-3x^2)^2 \cdot (-6x) = -18x(1-3x^2)^2 \).
4Step 4: Differentiate the Denominator using the Chain Rule
Let \( v = (3-x^2)^2 \). Again, use the chain rule: \( \frac{dv}{dx} = 2(3-x^2) \cdot \frac{d}{dx}(3-x^2) = 2(3-x^2) \cdot (-2x) = -4x(3-x^2) \).
5Step 5: Combine Results in the Quotient Rule Formula
Substitute \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) into the quotient rule: \[ \frac{d}{dx}f(x) = \frac{(3-x^2)^2(-18x(1-3x^2)^2) - (1-3x^2)^3(-4x(3-x^2))}{((3-x^2)^2)^2} \].
6Step 6: Simplify the Expression
Distribute and simplify each term in the numerator: \[ -18x(3-x^2)^2(1-3x^2)^2 + 4x(1-3x^2)^3(3-x^2) \]. Then simplify the fraction, continuing with any algebraic reduction necessary to explore cancelation or factoring benefits.
Key Concepts
Quotient RuleChain RuleSimplifying Fractions
Quotient Rule
The quotient rule is a fundamental technique in calculus differentiation that helps us determine the derivative of a function that is expressed as a quotient of two other functions. When a function is written in the form \( \frac{u(x)}{v(x)} \), the quotient rule states that its derivative is given by the formula:
- \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \)
Chain Rule
The chain rule is an essential tool in calculus used when dealing with compositions of functions. It's particularly handy when you have a function inside another function, as seen in expressions like \( (1-3x^2)^3 \). The chain rule allows us to take the derivative of such composite functions. To find the derivative using the chain rule, follow these steps:
- Differentiate the outer function, keeping the inner function unchanged.
- Multiply this result by the derivative of the inner function.
- Differentiating the outside, \( 3(1-3x^2)^2 \), and then multiplying by the derivative of the inside, \(-6x\), gives \( \frac{du}{dx} = -18x(1-3x^2)^2 \).
- For \( v = (3-x^2)^2 \), a similar process yields \( \frac{dv}{dx} = -4x(3-x^2) \).
Simplifying Fractions
Simplifying fractions is the process of reducing fractions to their simplest form. In calculus, particularly when applying rules like the quotient and chain rules, you often end up with complex fractions that are daunting at first glance. The goal is to simplify these results without losing the correct relationship or information of the expression. In our solution process, after applying the quotient rule, we reach the fractional expression: \[ \frac{-18x(3-x^2)^2(1-3x^2)^2 + 4x(1-3x^2)^3(3-x^2)}{((3-x^2)^2)^2} \] To simplify this, distribute and combine like terms in the numerator while looking for cancelation opportunities.
- Factor out common terms where possible to reduce the fraction.
- Carefully simplify the powers and coefficients.
Other exercises in this chapter
Problem 12
Find \(f^{(n)}(x)\) and \(f^{(n+1)}(x)\) if \(f(x)=x^{n}\).
View solution Problem 12
(a) The curve with equation \(y^{2}=x^{2}-x^{4}\) is shaped like the numeral eight. Find \(\frac{d y}{d x}\) at \(\left(\frac{1}{2}, \frac{1}{4} \sqrt{3}\right)
View solution Problem 13
Calculate the linear approximation for \(f(x)\) : $$f(x) \approx f(a)+f^{\prime}(a)(x-a)$$ \(f(x)=\frac{2}{1+x}\) at \(a=1\)
View solution Problem 13
Find \(c\) so that \(f^{\prime}(c)=0 .\) \(f(x)=x^{2}+6 x+9\)
View solution