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In Exercises \(65-70,\) find the value of \((f \circ g)^{\prime}\) at the given value of \(x .\) $$f(u)=u^{5}+1, \quad u=g(x)=\sqrt{x}, \quad x=1$$

Short Answer

Expert verified
The value of \\((f \, \circ \, g)'(1)\\) is \\frac{5}{2}\\.

Step by step solution

01

Understand the composite function

We need to find the value of the derivative of the composition of functions \(f \, \circ \, g\), which means \(f(g(x))\). Here, \(f(u) = u^5 + 1\) and \(g(x) = \sqrt{x}\). So, the composite function is \(f(g(x)) = (\sqrt{x})^5 + 1\).
02

Differentiate the composite function using the Chain Rule

To differentiate the function \(f(g(x))\), we apply the Chain Rule.The Chain Rule states that if \(h(x) = f(g(x))\), then the derivative \(h'(x) = f'(g(x)) \cdot g'(x)\).First, find \(f'(u)\): \(f(u) = u^5 + 1\) gives \(f'(u) = 5u^4\).Then, find \(g'(x)\): \(g(x) = \sqrt{x} = x^{1/2}\) gives \(g'(x) = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}\).
03

Substitute and evaluate derivatives

Now substitute \(u = g(x) = \sqrt{x}\) to get \(f'(g(x)) = 5(\sqrt{x})^4\).So, \(f'(g(x)) = 5x^2\). Combine all parts using the Chain Rule: \[(f \, \circ \, g)'(x) = f'(g(x)) \cdot g'(x) = 5x^2 \cdot \frac{1}{2\sqrt{x}} \].
04

Simplify and evaluate at given x

Simplify the expression: \[(f \, \circ \, g)'(x) = \frac{5x^2}{2\sqrt{x}} = \frac{5x^2}{2x^{1/2}} = \frac{5x^{2-1/2}}{2} = \frac{5x^{3/2}}{2} \].Now evaluate at \(x = 1\): \[(f \, \circ \, g)'(1) = \frac{5(1)^{3/2}}{2} = \frac{5 \cdot 1}{2} = \frac{5}{2} \].

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Key Concepts

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

Composite Functions
A composite function is built by applying one function to the result of another function. It connects two functions in a way that the output of one function becomes the input of another. In the problem we are solving, the functions involved are:
  • Function \( f(u) = u^5 + 1 \)
  • Function \( g(x) = \sqrt{x} \)
Here, the composite function, represented as \( (f \circ g)(x) \), means we first calculate \( g(x) \) and then apply \( f \) to the result. For this example, the composite function becomes \( f(g(x)) = (\sqrt{x})^5 + 1 \).
Understanding composite functions allows you to merge two processes into a single expression, focusing on the overall transformation resulting from both functions acting together.
Derivatives
In calculus, a derivative represents the rate at which a quantity changes as its input changes. For any function, its derivative is another function, which can tell you how the original function behaves - like its slope or steepness at any given point. It is a fundamental concept for understanding curves and rates of change in mathematical analysis.
The derivative is commonly denoted by a prime symbol such as \( f'(x) \) for a function \( f \). In the context of our specific composite function \( (f \circ g)(x) \), the derivative gives us insight into how the output changes when there is a small change in \( x \).
By applying the Chain Rule, which is essential when working with composite functions, we combine the derivatives of each component function. This enables us to find the derivative of the whole composite. For our functions:
  • The derivative of \( f(u) = u^5 + 1 \) is \( f'(u) = 5u^4 \)
  • The derivative of \( g(x) = \sqrt{x} \) is \( g'(x) = \frac{1}{2\sqrt{x}} \)
This setup allows us to differentiate even the most complex-looking functions.
Function Composition
Function composition involves combining two or more functions where the output of one function becomes the input of the next. In this case, it's represented by \( f(g(x)) \), which effectively translates as "apply \( g \) first, then \( f \) to the result."
Consider the functions:
  • \( g(x) \) operates first, transforming \( x \) into \( \sqrt{x} \).
  • \( f \) then takes \( \sqrt{x} \) and raises it to the fifth power, adding 1.
The beauty of function composition lies in its ability to model real-world systems where processes are sequential. It's like preparing a dish - you follow a sequence of steps where each ingredient undergoes specific transformations.
In practice, function composition allows for efficient problem-solving as it reduces a complex sequence of operations into a single expression, simplifying analysis and calculations.

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