Problem 42 - University Calculus: Early Transcendentals

Question

Find the derivatives of the functions in Exercises \(23-50\). $$g(x)=\frac{\tan 3 x}{(x+7)^{4}}$$

Step-by-Step Solution

Answer

The derivative is \( \frac{3 \sec^2 3x (x+7)^4 - 4 \tan 3x (x+7)^3}{(x+7)^8} \).

Step 1: Identify the Quotient Rule

The function \( g(x) = \frac{\tan 3x}{(x+7)^4} \) is a quotient, which can be solved by using the quotient rule. The quotient rule states that if you have a function \( u(x)/v(x) \), its derivative is \( \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} \). Here, \( u(x) = \tan 3x \) and \( v(x) = (x+7)^4 \).

Step 2: Compute the Derivative of the Numerator

Find the derivative of \( u(x) = \tan 3x \). Use the chain rule here: the derivative of \( \tan 3x \) is \( 3 \sec^2 3x \). Thus, \( u'(x) = 3 \sec^2 3x \).

Step 3: Compute the Derivative of the Denominator

Find the derivative of \( v(x) = (x+7)^4 \). Using the power rule, the derivative is \( 4(x+7)^3 \). Thus, \( v'(x) = 4(x+7)^3 \).

Step 4: Apply the Quotient Rule

Substitute \( u(x) \), \( v(x) \), \( u'(x) \), and \( v'(x) \) into the quotient rule:\[\frac{d}{dx}\left( \frac{\tan 3x}{(x+7)^4} \right) = \frac{(3 \sec^2 3x) \cdot (x+7)^4 - (\tan 3x) \cdot 4(x+7)^3}{((x+7)^4)^2}\]

Step 5: Simplify the Expression

Simplify the expression obtained in Step 4:Factor out common terms in the numerator if possible, and simplify the denominator, resulting in:\[\frac{3 \sec^2 3x \cdot (x+7)^4 - 4 \tan 3x \cdot (x+7)^3}{(x+7)^8}\]

Key Concepts

Quotient Rule

When dealing with derivatives of functions that are presented as a fraction, the Quotient Rule is your go-to tool. It is really made to handle these kinds of expressions efficiently. The rule states that for a function given by the quotient \( \frac{u(x)}{v(x)} \), the derivative is given by:

• \( \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} \)

The main idea is to differentiate the top function \( u(x) \) and the bottom function \( v(x) \) separately while keeping in mind that the denominator requires squaring at the end. The terms are subtracted in the numerator which is crucial as it impacts the result's sign. Always remember to first identify the two parts of the fraction as \( u(x) \) and \( v(x) \), then find their corresponding derivatives \( u'(x) \) and \( v'(x) \) before substitution.

Chain Rule

The Chain Rule comes into play when dealing with composite functions, that is, functions inside other functions. If you have a function \( g(x) = f(h(x)) \), the Chain Rule helps in finding its derivative efficiently. Its formula can be presented as:

• \( g'(x) = f'(h(x)) \times h'(x) \)

In essence, you first find the derivative of the outer function \( f \) evaluated at the inner function \( h(x) \), and then multiply it by the derivative of the inner function \( h(x) \). It's like peeling an onion, working from the outside layer to the inside. In our exercise, this method was used to differentiate \( \tan 3x \); since \( \tan \) is a function that starts with \( 3x \), making it necessary to apply the Chain Rule to achieve an accurate derivative.

Power Rule

The Power Rule is one of the most straightforward rules in calculus and is used when finding the derivative of a power of \( x \). It states that for a function \( f(x) = x^n \), where \( n \) is any real number:

• \( f'(x) = n \cdot x^{n-1} \)

It simply means you multiply by the current power and then subtract one from that power. This rule is particularly helpful when you have polynomial expressions. In this problem, it was applied to \( v(x) = (x+7)^4 \), signaling that with a slight modification, such as a change in the base to include \( x+7 \), the Power Rule still holds true, resulting in the derivative \( 4(x+7)^3 \). This simplified the process of calculating the derivative of the polynomial function involved.

Trigonometric Functions

In calculus, understanding the derivatives of trigonometric functions is crucial as these appear frequently in various problems. The basic derivatives that one should remember include:

• \( \frac{d}{dx} (\sin x) = \cos x \)
• \( \frac{d}{dx} (\cos x) = -\sin x \)
• \( \frac{d}{dx} (\tan x) = \sec^2 x \)

These derivatives are really handy when trig functions are a part of a more complex function. In this exercise, the derivative of \( \tan 3x \) was calculated. Notice that the derivative of \( \tan x \) is \( \sec^2 x \), and because of the Chain Rule, the factor \( 3 \) came into play, resulting in \( 3 \sec^2 3x \). It’s important to always be mindful of not only what the basic derivatives are but how they can interact with other rules like the Chain Rule to modify the results.