Problem 83 - University Calculus: Early Transcendentals

Question

In Exercises \(79-84,\) find the value of \((f \circ g)^{\prime}\) at the given value of \(x\). $$f(u)=\frac{2 u}{u^{2}+1}, u=g(x)=10 x^{2}+x+1, x=0$$

Step-by-Step Solution

Answer

The value of \((f \circ g)^{\prime}(0)\) is 0.

Step 1: Understand the Composition of Functions

First, we need to define the composite function \( (f \circ g)(x) \). This means substituting \( g(x) = 10x^2 + x + 1 \) into \( f(u) = \frac{2u}{u^2 + 1} \). So, the composite function \( (f \circ g)(x) \) becomes \( f(g(x)) = \frac{2(10x^2 + x + 1)}{(10x^2 + x + 1)^2 + 1} \).

Step 2: Differentiate the Composite Function Using the Chain Rule

To find the derivative \( (f \circ g)^{\prime}(x) \), we apply the chain rule. The chain rule states that \( (f \circ g)^{\prime}(x) = f^{\prime}(g(x)) \cdot g^{\prime}(x) \). We need to differentiate \( f(u) \) with respect to \( u \) and \( g(x) \) with respect to \( x \).

Step 3: Differentiate f(u) with respect to u

Using the quotient rule, differentiate \( f(u) = \frac{2u}{u^2 + 1} \). The quotient rule is \( f^{\prime}(u) = \frac{v \cdot u' - u \cdot v'}{v^2} \) for \( y = \frac{u}{v} \).Here \( u = 2u \) and \( v = u^2 + 1 \). Apply the quotient rule:\[ f^{\prime}(u) = \frac{(u^2 + 1) \cdot 2 - 2u \cdot 2u}{(u^2 + 1)^2} = \frac{2(u^2 + 1 - 2u^2)}{(u^2 + 1)^2} = \frac{2 - 2u^2}{(u^2 + 1)^2} \].

Step 4: Differentiate g(x) with respect to x

The function \( g(x) = 10x^2 + x + 1 \) is a simple polynomial. Its derivative is straightforward: \( g^{\prime}(x) = 20x + 1 \).

Step 5: Evaluate the Derivative at x = 0

Now that we have both derivatives, plug \( x = 0 \) into each derivative equation:- For \( f^{\prime}(g(x)) \), substitute \( g(0) = 1 \), so \[ f^{\prime}(1) = \frac{2 - 2 \cdot 1^2}{(1^2 + 1)^2} = \frac{0}{4} = 0 \].- \( g^{\prime}(0) = 20 \cdot 0 + 1 = 1 \).Thus, \( (f \circ g)^{\prime}(0) = 0 \times 1 = 0 \).

Key Concepts

Quotient Rule

The quotient rule is a method used to differentiate functions that are divided by one another. Suppose you have a function given by the division of two other functions, specifically, if you have a function expressed as \( y = \frac{u}{v} \), where both \(u\) and \(v\) are functions of \(x\). To differentiate \(y\) with respect to \(x\), you would apply the quotient rule, which states:

• \( f^{\prime}(x) = \frac{v \cdot u' - u \cdot v'}{v^2} \)

Here, \(u'\) is the derivative of \(u\) with respect to \(x\), and \(v'\) is the derivative of \(v\) with respect to \(x\).
In the exercise, the function \(f(u) = \frac{2u}{u^2 + 1}\) is a quotient of \(2u\) and \(u^2 + 1\). By applying the quotient rule, you differentiate the numerator and the denominator separately:

• The derivative of the numerator \(2u\) is \(2\).
• The derivative of the denominator \(u^2 + 1\) is \(2u\).

Thus, you combine these derivatives using the quotient rule formula to find \( f^{\prime}(u) = \frac{2(u^2 + 1) - 2u \cdot 2u}{(u^2 + 1)^2}\). This expression can be simplified to \( f^{\prime}(u) = \frac{2 - 2u^2}{(u^2 + 1)^2}\). Understanding how to apply the quotient rule is crucial when dealing with divisions in differentiation.

Composite Functions

Composite functions involve combining two functions where the output of one function becomes the input of another. It is written as \((f \circ g)(x) = f(g(x))\). In this problem, we have two functions, \(f(u)\) and \(g(x)\). These functions are combined to form a composite: \((f \circ g)(x)\).
The exercise involves determining the composite function by plugging \(g(x)\) into \(f(u)\). Given \(f(u) = \frac{2u}{u^2 + 1}\) and \(u = g(x) = 10x^2 + x + 1\), substituting \(g(x)\) into \(f(u)\) yields the composite function \(f(g(x)) = \frac{2(10x^2 + x + 1)}{(10x^2 + x + 1)^2 + 1}\).
This shows that the complexity of the composite function depends on the complexity of \(g(x)\) and \(f(u)\). By breaking it down into steps—first adjusting the inner function and then adjusting the outer function—we can better manage and understand the transformation process involved in making composite functions.

Polynomial Differentiation

Polynomial differentiation is a straightforward technique used in calculus to find the rate at which a polynomial changes. Polynomials are algebraic expressions that include coefficients and variables raised to various powers. The process of differentiating a polynomial involves applying the power rule.
The power rule states that if you have a term like \(ax^n\), its derivative is given by \(n \cdot ax^{n-1}\). This rule greatly simplifies finding derivatives of polynomial functions.

• For instance, consider the function \(g(x) = 10x^2 + x + 1\).

To differentiate \(g(x)\), apply the power rule to each term:

• The derivative of \(10x^2\) is \(20x\), because \(2 \cdot 10x^{2-1} = 20x\).
• The derivative of \(x\) is \(1\), since \(1x^{1-1} = 1\).
• The derivative of a constant \(1\) is \(0\), as constants do not change.

Aggregating these, the derivative \(g^{\prime}(x)\) is \(20x + 1\). This method is efficient and allows us to quickly find how polynomials change, which is an essential part of solving many calculus problems, including composite function derivations.