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Use the Chain Rule to differentiate each function. You may need to apply the rule more than once. $$ f(x)=\left(2 x^{3}+(4 x-5)^{2}\right)^{6} $$

Short Answer

Expert verified
The derivative is \( 6(2x^3 + (4x-5)^2)^5(6x^2 + 8(4x-5)) \).

Step by step solution

01

Identify the Outer and Inner Functions

The function given is \( f(x) = ig(2x^3 + (4x-5)^2\big)^6 \). The outer function is \( u^6 \), where \( u = 2x^3 + (4x-5)^2 \). The inner function is \( v^2 \), where \( v = 4x-5 \). We will need to differentiate step by step, using the chain rule on each part.
02

Differentiate the Outer Function

Differentiate the outer function with respect to \( u \). The derivative of \( u^6 \) with respect to \( u \) is \( 6u^5 \). So, \( \frac{d}{du}[u^6] = 6u^5 \).
03

Differentiate the Inner Function \( u \)

The inner function is \( u = 2x^3 + (4x-5)^2 \). Use the sum rule to differentiate: \( \frac{d}{dx}[2x^3] = 6x^2 \) and \( \frac{d}{dx}[(4x-5)^2] = 2(4x-5)(4) \). So, the derivative of \( u \) with respect to \( x \) is \( 6x^2 + 8(4x-5) \).
04

Differentiate the Inner Function \( v \)

To differentiate \( (4x-5)^2 \) more completely, find the derivative of \( v = 4x-5 \). We have \( \frac{d}{dx}[4x-5] = 4 \). The result was already included in \( 8(4x-5) \) as part of differentiating \( (4x-5)^2 \).
05

Apply the Chain Rule

Combine all parts according to the chain rule. We have \( \frac{df}{dx} = \frac{d}{du}[u^6] \cdot \frac{du}{dx} \). From previous steps, \( \frac{d}{du}[u^6] = 6u^5 \) and \( \frac{du}{dx} = 6x^2 + 8(4x-5) \). Therefore, \( \frac{df}{dx} = 6u^5(6x^2 + 8(4x-5)) \).
06

Substitute Back \( u \)

Substitute \( u = 2x^3 + (4x-5)^2 \) back into the expression for \( \frac{df}{dx} \). Thus, the derivative \( \frac{df}{dx} = 6(2x^3 + (4x-5)^2)^5(6x^2 + 8(4x-5)) \).

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Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus, which focuses on finding the rate at which a function changes at any point. When you differentiate a function, you determine its derivative—this tells you how quickly or slowly the function's value is increasing or decreasing. In the given exercise, we're dealing with a composite function, which requires the use of the chain rule for differentiation. The chain rule helps us take the derivative of a function nested inside another function. As we proceed with differentiation, we often encounter basic derivative rules such as the power rule, the sum rule, and the product rule. These rules make the process of finding derivatives more systematic and manageable.
Outer Function
In composite functions, the outer function is the primary function that wraps around the rest of the expression. For our exercise, the outer function is represented as
  • \(u^6\)
Here, the whole expression
  • \(2x^3 + (4x-5)^2\)
is the inner part that gets raised to the 6th power. When differentiating the outer function, you're focusing on its structure, ignoring the complexity of what lies inside. The derivative of the function
  • \(u^6\)
with respect to
  • \(u\)
becomes
  • \(6u^5\).
This step simplifies the process as it allows us to break down complex expressions into manageable parts.
Inner Function
In a composite function, identifying the inner function is crucial for applying the chain rule effectively. The inner function is the expression inside the outer function, which can itself be complex. In our example, the inner function
  • \( u = 2x^3 + (4x-5)^2\)
combines a polynomial and another function nested within. To differentiate, we apply basic differentiation rules like:
  • The derivative of \(2x^3\) is \(6x^2\).
  • For \((4x-5)^2\), we use the chain rule again: first, differentiate the square, yielding \(2(4x-5)\), then multiply by the derivative of \(4x-5\), which is \(4\), resulting in \(8(4x-5)\).
Together, these derivatives combine to form the full derivative of the inner function, \(6x^2 + 8(4x-5)\).
Expression Simplification
Expression simplification is a key step in finalizing the differentiation process. After finding the derivatives of both the outer and inner functions, you need to combine them. By applying the chain rule, you multiply the derivative of the outer function by the derivative of the inner function. In our example, the chain rule gives us
  • \(\frac{df}{dx} = 6u^5(6x^2 + 8(4x-5))\).
Subsequently, it's important to substitute back the value of
  • \(u\)
, resulting in:
  • \( \frac{df}{dx} = 6(2x^3 + (4x-5)^2)^5(6x^2 + 8(4x-5))\).
Simplifying expressions helps to provide a clearer and more concise representation of the derivative, making it easier to understand and apply in further calculations.

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Most popular questions from this chapter

Then estimate the \(x\) -values at which tangent lines are horizontal. $$ f(x)=x^{4}-3 x^{2}+1 $$

It has been shown that the home range, in hectares, of a carnivorous mammal weighing \(w\) grams can be approximated by $$ H(w)=0.11 w^{1.36} $$ (Source: Based on information in Emlen, J. M., Ecology: An Evolutionary Approach, p. \(216,\) Reading, MA: Addison-Wesley, 1973 ; and Harestad, A. S., and Bunnel, F. L., "Home Range and Body Weight-A Reevaluation," Ecology, Vol. 60, No. 2, pp. 405-418.) a) Find the average rate at which a carnivorous mammal's home range increases as the animal's weight grows from \(500 \mathrm{~g}\) to \(700 \mathrm{~g}\). b) Find \(\frac{H(300)-H(200)}{300-200}\). What does this rate represent?

Suppose that in \(t\) hours, a truck travels \(s(t)\) miles, where $$ s(t)=10 t^{2} $$ a) Find \(s(5)-s(2)\). What does this represent? b) Find the average rate of change of distance with respect to time as \(t\) changes from \(t_{1}=2\) to \(t_{2}=5\). This is also average velocity.

Let \(f\) and \(g\) be differentiable over an open interval containing \(x=a\). If $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{0}{0} \quad \text { or } \quad \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{\pm \infty}{\pm \infty} $$ and if \(\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right)\) exists, then $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right) $$ The forms \(0 / 0\) and \(\pm \infty / \pm \infty\) are said to be indeterminate. In such cases, the limit may exist, and l'Hôpital's Rule offers a way to find the limit using differentiation. For example, in Example 1 of Section \(1.1,\) we showed that $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=2 $$ Since, for \(x=1,\) we have \(\left(x^{2}-1\right)(x-1)=0 / 0,\) we differentiate the numerator and denominator separately, and reevaluate the limit: $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=\lim _{x \rightarrow 1}\left(\frac{2 x}{1}\right)=2 $$ Use this method to find the following limits. Be sure to check that the initial substitution results in an indeterminate form. $$ \lim _{x \rightarrow \infty}\left(\frac{4 x^{2}+x-3}{2 x^{2}+1}\right) $$

For each function, find the points on the graph at which the tangent line is horizontal. If none exist, state that fact. $$ y=-x^{3}+1 $$

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