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Find the derivative of the function. \(f(x)=\frac{x^{2}}{\left(x^{2}-1\right)^{4}}\)

Short Answer

Expert verified
The short answer is: \(f'(x) = \frac{x(x^2 - 1)^3(-6x^2 + 2)}{(x^2 - 1)^8}\)

Step by step solution

01

Identify the functions u(x) and v(x)

Here we have: \(u(x) = x^2\) \(v(x) = (x^2 - 1)^4\)
02

Find the derivatives of u(x) and v(x)

Using the power rule and the chain rule, we will differentiate u(x) and v(x) as follows. Power rule states: \(\frac{d}{dx}(x^n) = nx^{n-1}\) Chain rule states: \(\frac{d}{dx}(g(f(x)) = g'(f(x)) \cdot f'(x)\) Differentiate u(x): \(u'(x) = \frac{d}{dx}(x^2) = 2x\) Differentiate v(x): Let \(w(x) = x^2 - 1\), then \(v(x) = w(x)^4\) Apply chain rule: \(v'(x) = \frac{d}{dx}(w(x)^4) = 4w(x)^3 \cdot w'(x)\) Now differentiate w(x): \(w'(x) = \frac{d}{dx}(x^2 - 1) = 2x\) Now substitute back into the expression for \(v'(x)\): \(v'(x) = 4(x^2 - 1)^3 \cdot 2x = 8x(x^2 - 1)^3\)
03

Apply the quotient rule to find the derivative of the function

Using the quotient rule, we will now find the derivative of the given function: \(\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\) Substitute the values calculated in previous steps: \(\frac{d}{dx}\left(\frac{x^2}{(x^2 - 1)^4}\right) = \frac{(2x)(x^2 - 1)^4 - (x^2)(8x(x^2 - 1)^3)}{((x^2 - 1)^4)^2}\) Now, simplify the expression: \(\frac{d}{dx}\left(\frac{x^2}{(x^2 - 1)^4}\right) = \frac{2x(x^2 - 1)^4 - 8x^3(x^2 - 1)^3}{(x^2 - 1)^8}\) \(\frac{d}{dx}\left(\frac{x^2}{(x^2 - 1)^4}\right) = \frac{x(x^2 - 1)^3[2(x^2 - 1) - 8x^2]}{(x^2 - 1)^8}\) So the derivative of the given function is: \(\frac{d}{dx}\left(\frac{x^2}{(x^2 - 1)^4}\right) = \frac{x(x^2 - 1)^3(-6x^2 + 2)}{(x^2 - 1)^8}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Rule
When learning how to find derivatives, one of the first rules introduced is the Power Rule. This rule is quite straightforward and serves as a foundation in differentiation. It states that if you have a function in the form of \(x^n\), where \(n\) is any real number, the derivative of that function is calculated by multiplying \(n\) with \(x\) raised to the power of \(n-1\).
For instance, if you have \(x^2\), applying the power rule gives you the derivative \(2x\). This is because you bring down the exponent 2 as a coefficient and then subtract 1 from the exponent. This simple rule helps transform functions making them easier to analyze in calculus.
In our original exercise, the power rule was critical to find the derivative of \(u(x) = x^2\), simplifying the process by applying this basic principle.
Chain Rule
The Chain Rule is a method in calculus for differentiating compositions of functions. Think of it like a peeling an onion; you take it layer by layer. It’s especially useful when dealing with nested functions where one function is inside another.
This rule operates by taking the derivative of the outer function and multiplying it by the derivative of the inner function. Mathematically, if you have a function \( g(f(x)) \), the derivative is given by \( g'(f(x)) \cdot f'(x) \).
In the provided solution, the chain rule was applied to find the derivative of \( v(x) = (x^2 - 1)^4 \). First, identify the outer function \( w(x)^4 \) and the inner function \( w(x) = x^2 - 1 \). Then, differentiate both separately: differentiate the outer using the power rule, resulting in \( 4w(x)^3 \), and differentiate \( w(x) \) giving \( 2x \). Finally, multiply them together to obtain \( 8x(x^2 - 1)^3 \). The chain rule makes tackling complex problems like these manageable.
Quotient Rule
The Quotient Rule is an essential tool for differentiating functions that can be expressed as one function divided by another. This rule allows you to find the derivative of a quotient efficiently without needing to split it up or simplify it unnecessarily.
  • To apply the Quotient Rule, if you have functions \( u(x) \) and \( v(x) \), and you want to find the derivative of \( \frac{u(x)}{v(x)} \), use the formula:
\[ \frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \]
In the exercise, we used this rule to differentiate \( \frac{x^2}{(x^2 - 1)^4} \). This involves a few steps: first, identify \( u(x) \) and \( v(x) \) along with their derivatives as done in the previous steps. Then plug them into the quotient rule formula. After substituting:
  • \( u'(x) = 2x \)
  • \( v(x) = (x^2 - 1)^4 \)
  • \( v'(x) = 8x(x^2 - 1)^3 \)
The application leads to a simplified function involving both original and derived terms of \( u(x) \) and \( v(x) \).
By systematically applying the quotient rule, complex rational expressions become much simpler to handle and differentiate.

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Most popular questions from this chapter

HoTEL OccuPANCY RATES The occupancy rate of the allsuite Wonderland Hotel, located near an amusement park, is given by the function $$ r(t)=\frac{10}{81} t^{3}-\frac{10}{3} t^{2}+\frac{200}{9} t+60 \quad(0 \leq t \leq 12) $$ where \(t\) is measured in months, with \(t=0\) corresponding to the beginning of January. Management has estimated that the monthly revenue (in thousands of dollars/month) is approximated by the function $$ R(r)=-\frac{3}{5000} r^{3}+\frac{9}{50} r^{2} \quad(0 \leq r \leq 100) $$ where \(r\) is the occupancy rate. a. Find an expression that gives the rate of change of Wonderland's occupancy rate with respect to time. b. Find an expression that gives the rate of change of Wonderland's monthly revenue with respect to the occupancy rate. c. What is the rate of change of Wonderland's monthly revenue with respect to time at the beginning of January? At the beginning of July? Hint: Use the chain rule to find \(R^{\prime}(r(0)) r^{\prime}(0)\) and \(R^{\prime}(r(6)) r^{\prime}(6)\).

The total worldwide box-office receipts for a long-running movie are approximated by the function $$ T(x)=\frac{120 x^{2}}{x^{2}+4} $$ where \(T(x)\) is measured in millions of dollars and \(x\) is the number of years since the movie's release. How fast are the total receipts changing \(1 \mathrm{yr}, 3 \mathrm{yr}\), and \(5 \mathrm{yr}\) after its release?

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OIL SPILLS In calm waters, the oil spilling from the ruptured hull of a grounded tanker spreads in all directions. Assuming that the area polluted is a circle and that its radius is increasing at a rate of \(2 \mathrm{ft} / \mathrm{sec}\), determine how fast the area is increasing when the radius of the circle is \(40 \mathrm{ft}\).

Find the derivative of each function. \(h(t)=t^{2}(3 t+4)^{3}\)

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