Problem 26
Question
Use the Chain Rule-Power Rule to differentiate the given expression with respect to \(x\). $$ \left(\frac{x}{x+1}\right)^{4} $$
Step-by-Step Solution
Verified Answer
The derivative is \( \frac{4x^3}{(x+1)^5} \).
1Step 1: Identify the Function Structure
The given function is \( f(x) = \left(\frac{x}{x+1}\right)^4 \). Here, the inside function (u) is \( u = \frac{x}{x+1} \) and the outside function (v) is \( v = u^4 \). We need to apply the chain rule and power rule to differentiate this expression.
2Step 2: Differentiate the Outside Function using the Power Rule
Using the power rule, the derivative of \( u^n \) is \( n \cdot u^{n-1} \). Therefore, the derivative of \( v = u^4 \) with respect to \( u \) is \( 4u^3 \).
3Step 3: Differentiate the Inside Function
The inside function is \( u = \frac{x}{x+1} \). To differentiate it, we use the quotient rule, which states that \( \left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2} \). Here, \( f = x \) and \( g = x+1 \).
4Step 4: Apply the Quotient Rule
Differentiating \( f \) and \( g \): \( f' = 1 \) and \( g' = 1 \). Thus, \( u' = \frac{1(x+1) - x(1)}{(x+1)^2} = \frac{1}{(x+1)^2} \).
5Step 5: Combine Using the Chain Rule
The chain rule states that the derivative of a composite function \( v(u(x)) \) is \( \frac{dv}{du} \cdot \frac{du}{dx} \). Thus, \( \frac{df}{dx} = 4u^3 \cdot \frac{1}{(x+1)^2} \).
6Step 6: Substitute Back the Inside Function
Replace \( u \) with \( \frac{x}{x+1} \) back into the expression: \( \frac{df}{dx} = 4\left(\frac{x}{x+1}\right)^3 \cdot \frac{1}{(x+1)^2} \).
7Step 7: Simplify the Expression
The expression becomes \( \frac{4 x^3}{(x+1)^5} \) after simplifying. This is the derivative of the given function.
Key Concepts
Power RuleQuotient RuleDifferentiationComposite Function
Power Rule
The power rule is a fundamental rule in calculus, used for finding the derivative of functions of the form \( u^n \), where \( u \) is any function of \( x \), and \( n \) is a constant. In simple terms, when you have a function raised to a power, you multiply by that power and reduce the exponent by one.
For example, if we have \( v = u^4 \), applying the power rule gives:
For example, if we have \( v = u^4 \), applying the power rule gives:
- Bring down the exponent: Multiply by the power 4.
- Reduce the exponent by 1: Change \( u^4 \) to \( u^3 \).
Quotient Rule
The quotient rule is essential when differentiating functions given as fractions, where both the numerator and the denominator are functions of \( x \). The rule formula is:
\( \left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2} \), where \( f \) and \( g \) are functions of \( x \).
When differentiating the inside function \( u = \frac{x}{x+1} \), we take:
\( \frac{1(x+1) - x(1)}{(x+1)^2} = \frac{1}{(x+1)^2} \). This result represents \( \frac{du}{dx} \) and is critical in applying the chain rule subsequently.
\( \left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2} \), where \( f \) and \( g \) are functions of \( x \).
When differentiating the inside function \( u = \frac{x}{x+1} \), we take:
- \( f = x \) with \( f' = 1 \)
- \( g = x+1 \) with \( g' = 1 \)
\( \frac{1(x+1) - x(1)}{(x+1)^2} = \frac{1}{(x+1)^2} \). This result represents \( \frac{du}{dx} \) and is critical in applying the chain rule subsequently.
Differentiation
Differentiation is one of the core operations in calculus, providing a way to find the rate of change of a function. It helps us understand how a function behaves locally around a point. In this context, differentiation allows us to find the derivative of a function step by step, using various rules such as the power rule and quotient rule.
The process involves:
The process involves:
- Identifying the type of function (e.g., power, product, quotient).
- Choosing the appropriate differentiation rules.
- Applying these rules systematically to obtain the derivative.
Composite Function
A composite function is formed when one function is applied to the result of another function. This layered approach requires careful differentiation using the chain rule. The given function \( f(x) = \left(\frac{x}{x+1}\right)^4 \) is a composite function with:
The chain rule for differentiating composite functions says to first differentiate the outer function (applying the power rule), then multiply by the derivative of the inner function (using the quotient rule). This method ensures we fully capture how both layers contribute to the rate of change of the entire expression. The last steps involve substituting back for \( u \) and simplifying to find \( \frac{df}{dx} = \frac{4x^3}{(x+1)^5} \). Understanding each layer of the composite function simplifies the differentiation process.
- An outer function \( v = u^4 \)
- An inner function \( u = \frac{x}{x+1} \)
The chain rule for differentiating composite functions says to first differentiate the outer function (applying the power rule), then multiply by the derivative of the inner function (using the quotient rule). This method ensures we fully capture how both layers contribute to the rate of change of the entire expression. The last steps involve substituting back for \( u \) and simplifying to find \( \frac{df}{dx} = \frac{4x^3}{(x+1)^5} \). Understanding each layer of the composite function simplifies the differentiation process.
Other exercises in this chapter
Problem 26
Use implicit differentiation to find the normal line to the given curve at the given point \(P_{0}\). \(x^{4}-8 y+y^{6}=9 \quad P_{0}=(2,1)\)
View solution Problem 26
An expression for \(f(x)\) is given. Compute the first, second, and third derivatives of \(f(x)\) with respect to \(x\). \(x \log _{2}(x)\)
View solution Problem 26
Use the Inverse Function Derivative Rule to calculate \(\left(f^{-1}\right)^{\prime}(t)\). $$ f:(-\pi / 2, \pi / 2) \rightarrow(-\infty, \infty), f(s)=\tan (s)
View solution Problem 26
Use the Reciprocal Rule to compute the derivative of the given expression with respect to \(x\) $$ 1 /\left(x^{3}+x^{2}+1\right) $$
View solution