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Find the derivatives of the functions in Exercises \(19-40\) $$g(x)=\frac{\tan 3 x}{(x+7)^{4}}$$

Short Answer

Expert verified
Use the quotient rule on \(\frac{\tan 3x}{(x+7)^4}\), with final derivative formula given in Step 4.

Step by step solution

01

Identify the Rule to Use

The function given is a quotient of two functions: \(g(x) = \frac{\tan 3x}{(x+7)^4}\). To find the derivative, use the quotient rule: \(\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{v \cdot u' - u \cdot v'}{v^2}\). Identify \(u = \tan 3x\) and \(v = (x+7)^4\).
02

Differentiate the Numerator with the Chain Rule

Differentiate \(u = \tan 3x\). Using the chain rule, the derivative of \(\tan 3x\) is \(u' = \sec^2(3x) \cdot 3 = 3\sec^2(3x)\).
03

Differentiate the Denominator

Differentiate \(v = (x+7)^4\). Using the power rule, \(v' = 4(x+7)^3\).
04

Apply the Quotient Rule

Substitute \(u\), \(v\), \(u'\), and \(v'\) into the quotient rule formula: \[g'(x) = \frac{(x+7)^4 \cdot 3\sec^2(3x) - \tan 3x \cdot 4(x+7)^3}{((x+7)^4)^2}\].
05

Simplify the Expression

Simplify the numerator: \((x+7)^4 \cdot 3\sec^2(3x) - \tan 3x \cdot 4(x+7)^3\) and leave the denominator as \((x+7)^8\). The detailed simplification would require combining like terms in the numerator if possible.

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Key Concepts

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

Quotient Rule
When working with calculus derivatives, the quotient rule is an essential differentiation technique. It applies to functions that are the ratio of two other functions. For a function written as a quotient, say \( g(x) = \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), the quotient rule helps find \( g'(x) \), the derivative of \( g(x) \).
The formula is:
  • \( \frac{d}{dx}\left[\frac{u}{v}\right] = \frac{v \cdot u' - u \cdot v'}{v^2} \)
To apply the quotient rule, first differentiate the numerator function \( u \) to get \( u' \), and then differentiate the denominator function \( v \) to get \( v' \).
After computing these derivatives, substitute \( u \), \( u' \), \( v \), and \( v' \) back into the formula to find \( g'(x) \). By following these steps, you ensure that you correctly derive the expression for functions presented as fractions.
Chain Rule
Another fundamental tool in differentiation is the chain rule. It's particularly useful when dealing with composite functions, where one function is nested inside another.
The chain rule states if you have a composite function \( y = f(g(x)) \), the derivative \( \frac{dy}{dx} \) is found by multiplying the derivative of \( f \) with respect to \( g(x) \) by the derivative of \( g(x) \) with respect to \( x \), expressed as:
  • \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \)
In the given exercise, one instance of using the chain rule arises while differentiating \( \tan 3x \). Here, the function \( f(u) = \tan u \) with \( u = 3x \) results in \( f'(u) = \sec^2 u \).
Thus, applying the chain rule, \( \frac{d}{dx} \tan 3x = \sec^2(3x) \cdot 3 = 3\sec^2(3x) \). This showcases how the chain rule allows us to differentiate more complex function compositions by following a simple, systematic approach.
Differentiation Techniques
Differentiation techniques encompass various rules and methods used to find the derivative of functions. These include but are not limited to the power rule, product rule, quotient rule, and chain rule.
Each technique is suitable for particular types of functions:
  • Power Rule: Useful for functions of the form \( x^n \). Take \( \frac{d}{dx}(x^n) = nx^{n-1} \).
  • Product Rule: Used when differentiating the product of two functions, \( (fg)' = f'g + fg' \).
  • Quotient Rule: Essential for functions that are fractions, as discussed earlier.
  • Chain Rule: Helps with composite functions, as also covered above.
In the exercise at hand, these techniques come together to enable the differentiation of \( g(x)=\frac{\tan 3 x}{(x+7)^{4}} \).
The process begins by identifying which technique to apply, followed by precise differentiation. Each step systematically follows from the previous, resulting in the final derivative.
Understanding these techniques equips students with the ability to tackle a wide array of engineering and scientific problems effectively.

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Most popular questions from this chapter

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