Problem 66 - Calculus and its applications

Question

Let \(f(u)=2 u^{5}\) and \(g(x)=u=\frac{3-x}{4+x}\) Find \((f \circ g)^{\prime}(-10)\)

Step-by-Step Solution

Answer

The derivative \((f \circ g)'(-10)\) is approximately \(-42.85\).

Step 1: Identify the Composite Function

A composite function is formed when one function is applied to the result of another function. In this exercise, we are asked to find the derivative of the composite function \((f \circ g)(x)\). Here, \(f(u) = 2u^5\) and \(g(x) = \frac{3-x}{4+x}\). So the composite function is: \((f \circ g)(x) = f(g(x)) = f\left(\frac{3-x}{4+x}\right)\).

Step 2: Express the Composite Function

Substitute \(g(x) = \frac{3-x}{4+x}\) into \(f(u)\):\[(f \circ g)(x) = f\left(\frac{3-x}{4+x}\right) = 2 \left( \frac{3-x}{4+x} \right)^5\]Now \((f \circ g)(x)\) is \(2(\frac{3-x}{4+x})^5\).

Step 3: Find the Derivative of the Composite Function

Use the chain rule to find the derivative: \[(f \circ g)'(x) = \frac{d}{dx}\left[ 2 \left( \frac{3-x}{4+x} \right)^5 \right] = 2\cdot 5\left(\frac{3-x}{4+x}\right)^4 \cdot \frac{d}{dx}\left(\frac{3-x}{4+x}\right)\]First, find \(\frac{d}{dx}\left(\frac{3-x}{4+x}\right)\) using the quotient rule.

Step 4: Apply the Quotient Rule

The quotient rule for \(y = \frac{u}{v}\) is \(y' = \frac{u'v - uv'}{v^2}\). Let \(u = 3-x\) and \(v = 4+x\). Then:- \(u' = -1\)- \(v' = 1\)Substitute these into the quotient rule:\[\frac{dy}{dx} = \frac{(-1)(4+x) - (3-x)(1)}{(4+x)^2} = \frac{-4-x-(3-x)}{(4+x)^2} = \frac{-7}{(4+x)^2}\]

Step 5: Substitute Derivative Back into Chain Rule

Substitute \(\frac{dy}{dx} = \frac{-7}{(4+x)^2}\) into the expression from the chain rule result:\[(f \circ g)'(x) = 10\left(\frac{3-x}{4+x}\right)^4 \cdot \frac{-7}{(4+x)^2} = -70\left(\frac{3-x}{4+x}\right)^4/(4+x)^2\]

Step 6: Evaluate the Derivative at x = -10

Plug in \(x = -10\) into the expression we found:- Calculate \(\frac{3-(-10)}{4+(-10)} = \frac{13}{-6} = -\frac{13}{6}\)- Substitute into the expression:\[-70\left(-\frac{13}{6}\right)^4/(4+(-10))^2 = -70\left(-\frac{13}{6}\right)^4/36\]Subtract and simplify.

Step 7: Simplify the Expression

Calculate \( -70 \left(-\frac{13}{6}\right)^4/36 \):- Calculate \(\left(-\frac{13}{6}\right)^4 = \frac{28561}{1296}\)- Multiply: \(-70 \times \frac{28561}{1296}\times \frac{1}{36} = \frac{-1999270}{46656}\)- Simplify to \(-42.85\) approximately.

Key Concepts

Composite Functions

Composite functions occur when you combine two functions in such a way that the output of one function becomes the input of the other. Imagine this kind of operation as a two-step process. You start by evaluating a function and then immediately input that result into another function. In terms of notation, the composite function of two functions \( f \) and \( g \) is written as \( (f \, \circ \, g)(x) = f(g(x)) \). In the original exercise, \( f(u) = 2u^5 \) and \( g(x) = \frac{3-x}{4+x} \) are our two functions. To form the composite function \( (f \, \circ \, g)(x) \), we first evaluate \( g(x) \) and then use that result as input into \( f(u) \). This creates the expression \( f(g(x)) \), which is \( 2\left(\frac{3-x}{4+x}\right)^5 \). Helping you visualize this process can make understanding composite functions easier in calculus. Remember, the order of operations is crucial: follow the way they are nested.

Chain Rule

The chain rule is a fundamental technique in calculus used to differentiate composite functions. It helps us find the derivative of a function that is made up of other functions. If you have a composite function \( f(g(x)) \), the chain rule states that its derivative is \( f'(g(x)) \cdot g'(x) \). This means you first take the derivative of the outer function \( f \) evaluated at \( g(x) \), then multiply it by the derivative of the inner function \( g \). In our example, the outer function is \( f(u) = 2u^5 \). So, the derivative \( f'(u) \) is \( 10u^4 \). For the inner function \( g(x) = \frac{3-x}{4+x} \), we use the quotient rule to find \( g'(x) \). The chain rule simplifies the process of finding derivatives in situations involving nested functions, enabling you to tackle complex problems step-by-step.

Quotient Rule

The quotient rule is used when differentiating functions that are the quotient of two other functions. If you have a function \( y = \frac{u}{v} \), the derivative is given by \( y' = \frac{u'v - uv'}{v^2} \). It allows us to differentiate expressions with fractions smoothly. For the function \( g(x) = \frac{3-x}{4+x} \) from our exercise, we set \( u = 3-x \) and \( v = 4+x \). Calculating their derivatives gives \( u' = -1 \) and \( v' = 1 \). Embedding these into the quotient rule formula, we obtain \( \frac{dy}{dx} = \frac{-7}{(4+x)^2} \). The quotient rule links to the chain rule, as it provides the necessary derivative component for the inner function in composite function differentiation.

Derivatives

Derivatives are a cornerstone of calculus, representing rates of change. They tell us how a function's output value changes as its input value changes. When working with complex functions, being able to compute derivatives effectively helps us understand dynamic systems. In our example, the goal was to find the derivative of the composite function \( (f \, \circ \, g)(x) \) and evaluate it at \( x = -10 \). By simplifying the function further, we multiply \( 10\left(\frac{3-x}{4+x}\right)^4 \) by \( \frac{-7}{(4+x)^2} \) as calculated using the chain and quotient rules. After simplifying and plugging \( x = -10 \) into the derivative expression, further simplification yields the approximate value \( -42.85 \). Through understanding derivatives, you gain insights into how variables influence functions in dynamic situations and can predict future behavior of a function based on its current rate of change.