Chapter 2: Problem 9
Differentiate the following functions: \(u=x \sqrt[4]{a+b x+c x^{2}}\)
Short Answer
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Question: Find the derivative of the function u(x) = \(x \sqrt[4]{a+bx+cx^2}\).
Answer: The derivative of the given function is u'(x) = \((a+bx+cx^2)^{\frac{1}{4}} + x\frac{1}{4}(a+bx+cx^2)^{\frac{-3}{4}}(b+2cx)\).
Step by step solution
01
Identify all the functions involved
Here we have the outer function being a product of two functions: f(x) = x and g(x)=\(\sqrt[4]{a+bx+cx^2}\) which we will rewrite as g(x)=\((a+bx+cx^2)^\frac{1}{4}\) for easier differentiation.
02
Use the product rule to find the whole function's derivative
The product rule states that the derivative of a product of two functions is equal to the derivative of the first function multiplied by the second function plus the first function multiplied by the derivative of the second function, that is \((u(x)v(x))' = u'(x)v(x) + u(x)v'(x)\).
So in order to find the derivative of u(x), we need to first find the derivatives of f(x) and g(x).
03
Find the derivative of f(x)=x
By using the power rule, which states that \((x^n)' = nx^{n-1}\), we find that the derivative of x is 1, i.e., \(f'(x)=1\).
04
Find the derivative of g(x)=\((a+bx+cx^2)^{\frac{1}{4}}\)
We will employ the chain rule here, which states that \((h(g(x)))' = h'(g(x))g'(x)\).
Our function \((a+bx+cx^2)^{\frac{1}{4}}\) can be seen as the composition of two functions:
\(h(y)=y^{\frac{1}{4}}\) and \(g(x)=a+bx+cx^2\)
First, we'll find the derivative of h(y). By using the power rule, we'll get \(h'(y)=\frac{1}{4}y^\frac{-3}{4}\)
Furthermore, let's find the derivative of g(x). By using the power rule again, for g(x)=\(a+bx+cx^2\), we'll get \(g'(x)=b+2cx\)
Now, let's use the chain rule to find the derivative of g(x) = \((a + bx+ cx^2)^\frac{1}{4}\). By applying it, we get:
\(g'(x)=h'(g(x))g'(x)\)
\( = \frac{1}{4}(a+bx+cx^2)^{\frac{-3}{4}}(b+2cx)\)
05
Use the results to find the derivative of u(x)
Now that we have the derivatives of both f(x) and g(x), we can use the product rule to find the derivative of u(x):
u'(x) = f'(x)g(x) + f(x)g'(x)
\(=\left(1\right)\left((a+bx+cx^2)^{\frac{1}{4}}\right) + x\left(\frac{1}{4}(a+bx+cx^2)^{\frac{-3}{4}}(b+2cx)\right)\)
Simplifying it, we have:
u'(x)=\((a+bx+cx^2)^{\frac{1}{4}} + x\frac{1}{4}(a+bx+cx^2)^{\frac{-3}{4}}(b+2cx)\)
And that's the derivative of the given function.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding the Product Rule
The product rule is a fundamental concept in calculus, used for finding the derivative of a product of two functions. It states that if you have two functions, say \( f(x) \) and \( g(x) \), their derivative when multiplied is given by:
In this exercise, the function \( u(x) = x \sqrt[4]{a+bx+cx^{2}} \) is a product of \( f(x)=x \) and \( g(x)=\sqrt[4]{a+bx+cx^{2}} \).
Differentiating both \( f(x) \) and \( g(x) \) separately and applying the product rule makes it possible to find \( u'(x) \) effectively.
- \((f(x)g(x))' = f'(x)g(x) + f(x)g'(x)\)
In this exercise, the function \( u(x) = x \sqrt[4]{a+bx+cx^{2}} \) is a product of \( f(x)=x \) and \( g(x)=\sqrt[4]{a+bx+cx^{2}} \).
Differentiating both \( f(x) \) and \( g(x) \) separately and applying the product rule makes it possible to find \( u'(x) \) effectively.
The Power Rule Simplified
The power rule is one of the simplest and most frequently used rules for differentiation, especially for polynomials. It states that for a function \( x^n \), the derivative is:
This principle was used in the step where we found the derivative of \( f(x) = x \).
The power rule also comes into play when differentiating each monomial term in \( g(x)=a+bx+cx^2 \), helping us evaluate terms like \( b \) and \( 2cx \).
- \( n \cdot x^{n-1} \)
This principle was used in the step where we found the derivative of \( f(x) = x \).
The power rule also comes into play when differentiating each monomial term in \( g(x)=a+bx+cx^2 \), helping us evaluate terms like \( b \) and \( 2cx \).
Navigating the Chain Rule
The chain rule is essential for differentiating composite functions, those which involve one function inside another. It is written as:
The chain rule helps break down this complexity.
First, we found the derivative of \( h(y)\) using the power rule, leading to \( h'(y)=\frac{1}{4}y^{\frac{-3}{4}} \).
Next, we found the derivative \( g'(x) \) for \( g(x) \).
Applying the chain rule, we integrated these pieces to get the derivative of \( g(x) \). This illustrates how the chain rule allows differentiation of compound curves by taking into consideration the derivative of the outer and inner functions.
- \( (h(g(x)))' = h'(g(x)) \cdot g'(x) \)
The chain rule helps break down this complexity.
First, we found the derivative of \( h(y)\) using the power rule, leading to \( h'(y)=\frac{1}{4}y^{\frac{-3}{4}} \).
Next, we found the derivative \( g'(x) \) for \( g(x) \).
Applying the chain rule, we integrated these pieces to get the derivative of \( g(x) \). This illustrates how the chain rule allows differentiation of compound curves by taking into consideration the derivative of the outer and inner functions.