Question

In Exercises, find the derivative of the function. $$ f(x)=\frac{\left(e^{x}+e^{-x}\right)^{4}}{2} $$

Short answer

Hence, the derivative of the function \( y=\frac{\ln x}{x^{2}} \) is \( y'=\frac{1 - 2\ln x}{x^{3}} \)

Step-by-step solution

Identify the functions

In our given equation \( y=\frac{\ln x}{x^{2}} \), two functions can be identified. These are: \( f(x) = \ln x \) and \( g(x) = x^{2} \) respectively.

Find the derivative of the functions

To apply the quotient rule, we first need to find the derivatives of the functions identified in step 1. The derivative of \( f(x) = \ln x \) is \( f'(x) = \frac{1}{x} \). The derivative of \( g(x) = x^{2} \) is \( g'(x) = 2x \). These will be used in applying the quotient rule.

Apply the quotient rule

Now that we have \( f'(x) \) and \( g'(x) \), we can apply the quotient rule to determine the derivative of \( y \). Using the quotient rule as mentioned in the analysis: \( y'=\frac{f'(x)g(x) - f(x)g'(x)} {[g(x)]^{2}} \), substituting the corresponding values gives us \( y'=\frac{(\frac{1}{x} * x^{2}) - (\ln x * 2x)} {(x^{2})^{2}} \). After simplifying further, we get \( y'=\frac{1 - 2\ln x}{x^{3}} \)

Key concepts

Quotient Rule

Understanding the quotient rule is essential in calculus when faced with the task of differentiating functions that are divided by each other, a scenario that often stumps students. In essence, the quotient rule provides a systematic method for finding the derivative of a ratio of two functions. Let's say we have a function in the form of \( \frac{f(x)}{g(x)} \), where both \( f(x) \) and \( g(x) \) are differentiable functions. The quotient rule states that the derivative of this function, notated as \( \left( \frac{f(x)}{g(x)} \right)' \), is given by \( \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^{2}} \).

Here's a more digestible breakdown: start by differentiating the top function (\( f(x) \) - the numerator) and the bottom function (\( g(x) \) - the denominator) separately. The derivative of the entire fraction then involves taking these two derivatives and following the pattern of 'difference over square' — the difference of the product of the top function's derivative and the bottom function with the product of the top function and bottom function's derivative, all over the square of the bottom function. This compact formula does wonders, simplifying the process of differentiation when division is involved.

Exponential Functions Derivatives

Exponential functions often make an appearance in calculus problems due to their ubiquitous nature in describing growth and decay processes. The derivative of an exponential function can appear intimidating at first, but with the right approach, it becomes straightforward. The general form of an exponential function is \( e^{x} \), where \( e \) is Euler's number, approximately equal to 2.71828. This special constant has the unique property that the rate of change of the function \( e^{x} \) with respect to \( x \) is equal to the function itself, meaning \( \frac{d}{dx}e^{x} = e^{x} \).

In calculus problem solving, when you come across \( e^{x} \) raised to a power or multiplied by a coefficient, simply remember that the derivative will include that power or coefficient in the result. For instance, if the function is \( (e^{x})^{n} \), you'll need to apply the chain rule, another essential differentiation tool, in combination with the derivative of the exponential function. The key is to take these unique properties of \( e^{x} \) into account when solving calculus problems, allowing for accurate and confident computation of derivatives for exponential functions.

Calculus Problem Solving

Solving calculus problems efficiently requires familiarity with a variety of rules and techniques. The complexity increases when functions combine operations, such as multiplication, division, and exponentiation. In these scenarios, identify the different aspects of the function—like recognising the presence of a quotient (division) or the application of an exponential function—and then decide which calculus rules apply. In the provided textbook exercise, we need to find the derivative of \( f(x)=\frac{(e^{x}+e^{-x})^{4}}{2} \).

To tackle this problem, start by recognizing that the function inside the parentheses is an exponential function, and the entire expression is a quotient, suggesting that the chain rule and quotient rule will be useful tools here. The chain rule allows us to differentiate composite functions, while the quotient rule provides a clear path to handle the division. Memorizing these rules isn't enough; applying them effectively requires practice and understanding. Work out simpler problems to build up skills and confidence in identifying which rules to apply and when. Remember to write out all the steps clearly to avoid confusion, and double-check the algebra as many errors in calculus are actually algebra mistakes. An attentive and methodical approach combined with a solid grasp of the basic principles of differentiation will lead to successful problem solving in calculus.