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Calculate the derivatives of the functions in Exercises 1-46. HINT [See Example 1.] \(f(x)=\left[(6.4 x-1)^{2}+(5.4 x-2)^{3}\right]^{2}\)

Short Answer

Expert verified
The short answer is: \(f'(x)=2\left[(6.4x-1)^{2}+(5.4x-2)^{3}\right] \cdot \left[2(6.4x-1)(6.4) + 3(5.4x-2)^{2}(5.4)\right]\)

Step by step solution

01

Identify the functions involved

We are given the function: \(f(x)=\left[(6.4x-1)^{2}+(5.4x-2)^{3}\right]^{2}\) Let's rewrite this function as \(f(x) = (u(x)+v(x))^2\), where \(u(x)=(6.4x-1)^2\) and \(v(x)=(5.4x-2)^3\).
02

Differentiate the inner functions u(x) and v(x)

We will now differentiate the two inner functions \(u(x)\) and \(v(x)\). Using the power rule, which states that the derivative of \(x^n\) is \(nx^{n-1}\), we get: \[u'(x)=2(6.4x-1)^{1}(6.4)\] \[v'(x)=3(5.4x-2)^{2}(5.4)\]
03

Differentiate the outer function

Now, we will differentiate the outer function: \(f'(x)=2(u(x)+v(x))^{1}\)
04

Apply the chain rule

We will now apply the chain rule: \[f'(x)=2(u(x)+v(x))^{1} \cdot \left(u'(x)+v'(x)\right)\] Substitute the expressions of \(u(x)\), \(v(x)\), \(u'(x)\), and \(v'(x)\) obtained from previous steps: \[f'(x)=2\left[(6.4x-1)^{2}+(5.4x-2)^{3}\right] \cdot \left[2(6.4x-1)(6.4) + 3(5.4x-2)^{2}(5.4)\right]\] This is the derivative of the given function.

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Key Concepts

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

Power Rule
In calculus, the power rule is a straightforward and fundamental tool for differentiating functions. It applies to functions of the form \(x^n\) where \(n\) is any real number. The power rule states that the derivative of \(x^n\) is \(nx^{n-1}\). This means you multiply by the exponent \(n\) and then decrease the exponent by one. This rule simplifies finding derivatives because it turns potentially complex expressions into manageable forms.
To apply the power rule effectively, identify terms that follow the \(x^n\) structure. For example, in the original function \(u(x)=(6.4x-1)^2\), we see an exponent of 2. Using the power rule:
  • The derivative \(u'(x)\) becomes \(2(6.4x-1)^{1}\times 6.4\).
  • Essentially, multiply the original exponent by the inner function's derivative.
This quick conversion is particularly useful for polynomial expressions encountered in calculus.
Chain Rule
The chain rule is an essential derivative technique used when functions are nested within each other. Essentially, it allows you to differentiate composite functions systematically. The chain rule states that the derivative of a composite function \(f(g(x))\) is \(f'(g(x)) \cdot g'(x)\). Think of it as peeling layers of an onion, removing one layer at a time.
For the given function \(f(x)=\left[(6.4x-1)^{2}+(5.4x-2)^{3}\right]^{2}\), the chain rule comes into play because we have a function raised to a power, and those bases themselves are also composite functions. The steps:
  • Differentiate the outer function, leaving the inner functions intact initially.
  • Multiply by the derivative of the inner functions, which involves the power rule in our case.
So, the derivative of the outer function is \(2(u(x) + v(x))^{1}\), which is then multiplied by the inner derivatives \(u'(x)+v'(x)\), as shown in the step-by-step solution. This approach efficiently tackles complex nested functions.
Differentiation Techniques
Understanding differentiation techniques is crucial for solving calculus problems accurately and efficiently. Differentiation refers to the process of finding a function's derivative, which represents the rate of change of the function. In this exercise, a combination of techniques is involved:
  • **Power Rule**: Used for polynomial-like parts, as seen in finding \(u'(x)\) and \(v'(x)\).
  • **Chain Rule**: Used to differentiate composite functions, especially when layers of functions are nested.
By employing these techniques together, we can tackle more complicated expressions. Start by breaking down the function into its components and identifying which rules apply to each part. Ensure each layer is differentiated correctly using corresponding rules.
A systematic approach enhances understanding and efficiency, helping you apply the right steps in sequence for accurate results.

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Most popular questions from this chapter

In Exercises 55-60, compute the indicated derivative using the chain rule. HINT [See Quick Examples on page 828.] \(y=3 x-2 ; \frac{d x}{d y}\)

Calculate the derivatives of the functions in Exercises 1-46. HINT [See Example 1.] \(g(x)=\left(2 x^{2}+x+1\right)^{-3}\)

Oil Imports from Mexico Daily oil production in Mexico and daily U.S. oil imports from Mexico during 2000-2004 can be approximated by $$ \begin{array}{ll} P(t)=3.0+0.13 t \text { million barrels } & (0 \leq t \leq 4) \\ I(t)=1.4+0.06 t \text { million barrels } & (0 \leq t \leq 4) \end{array} $$ where \(t\) is time in years since the start of \(2000 .^{21}\) a. What are represented by the functions \(P(t)-I(t)\) and \(I(t) / P(t) ?\) b. Compute \(\left.\frac{d}{d t}\left[\frac{I(t)}{P(t)}\right]\right|_{t=3}\) to two significant digits. What does the answer tell you about oil imports from Mexico?

The following graph shows the approximate value of home prices and existing home sales in 2006–2010 as a percentage change from 2003, together with quadratic approximations The quadratic approximations are given by $$ \text { Home Prices: } \quad P(t)=t^{2}-10 t+41 \quad(0 \leq t \leq 4) $$ Existing Home Sales: \(\quad S(t)=1.5 t^{2}-11 t \quad(0 \leq t \leq 4)\) where \(t\) is time in years since the start of 2006 . Use the chain rule to estimate \(\left.\frac{d S}{d P}\right|_{t=2} .\) What does the answer tell you about home sales and prices? HINT [See Quick Example 2 on page 828.]

Calculate the derivatives of the functions in Exercises 1-46. HINT [See Example 1.] \(f(x)=\left(2 x^{2}-2\right)^{-1}\)

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