Question

In Exercises, find the derivative of the function. $$ y=x e^{x}-4 e^{-x} $$

Short answer

The derivative of the function \(y=\ln \left(x \sqrt{x^2-1}\right)\) is \(y'=\frac{1+x^2}{x\sqrt{x^2-1}}\).

Step-by-step solution

Identify the outer function and the inner function

Our outer function here is the natural logarithm: \(y=\ln(u)\), while our inner function is \(u=x \sqrt{x^2-1}\). We will use the chain rule to differentiate this nested function, which states that the derivative is obtained by differentiating the outer function and multiplying it by the derivative of the inner function.

Differentiate the outer function while keeping the inner function in place

The derivative of the natural logarithm \(y=\ln(u)\) is \(y'=\frac{1}{u}\). For our function, we write it as \(\frac{1}{x \sqrt{x^2-1}}\).

Differentiate the inner function using the product rule

The inner function \(u=x \sqrt{x^2-1}\) requires the product rule for differentiation. This rule states that the derivative of two multiplied functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. Here, we consider \(x\) as the first function and \(\sqrt{x^2-1}\) as the second function. We then differentiate both. Differentiating \(x\) gives \(1\), and for \(\sqrt{x^2-1}\) we use the chain rule again. Letting \(r=x^2-1\), we get \(r'=\frac{1}{2}(r)^{-1/2}(2x)\). Combining these with the product rule gives \(u'=1*\sqrt{x^2-1} + x*\frac{x}{\sqrt{x^2-1}}\).

Multiply the derived outer and inner functions

Finally, we multiply the results from step 2 and 3 together, following the chain rule. This gives our derivative: \(y'=\frac{1}{x \sqrt{x^2-1}}*(1*\sqrt{x^2-1} + x*\frac{x}{\sqrt{x^2-1}})\). The terms simplify to \(y'=\frac{1+x^2}{x\sqrt{x^2-1}}\).

Key concepts

Chain Rule

The chain rule is essential when dealing with composite functions. These are functions nested within each other, resembling a stack of boxes with one box sitting inside another. In mathematics, the chain rule helps us differentiate these stacked functions efficiently. It's like peeling an onion, layer by layer.
To apply the chain rule, first identify the outer and inner functions. Differentiate the outer function with respect to the inner function while keeping the inner function intact. Next, multiply this by the derivative of the inner function. For example, if you have a function of the form \(h(x) = f(g(x))\), then the derivative \(h'(x)\) is \(f'(g(x)) \, g'(x)\).

• Identify the layers: Know which function is on the outside and inside.
• Differentiate the outer function while keeping the inside untouched.
• Find the derivative of the inner function.
• Multiply the two derivatives together.

This step-by-step approach ensures accuracy in finding the derivative of complex, composite functions.

Product Rule

Understanding the product rule is crucial when differentiating functions that are products of two separate functions. Picture it like gears meshing together in a machine; each gear must work in tandem for the machine to function smoothly.
The product rule states that for two functions \(u(x)\) and \(v(x)\), the derivative of their product is \((uv)' = u'v + uv'\). This means that you differentiate each function one at a time while multiplying by the other, and then add the results together. Here's a simple way to remember it:

• Differentiate the first function, leave the second function as it is, and multiply them together.
• Add the result to the derivative of the second function, multiplying by the original first function.

When applying this to our specific problem involving \(u = x\) and \(v = \sqrt{x^2 - 1}\), remember each component of the function plays a role, just like every gear in a machine is necessary for the mechanism to work.

Logarithmic Differentiation

Logarithmic differentiation is a powerful tool, especially when dealing with functions involving products, powers, or complex expressions. It's like having a mathematical magnifying glass that makes complicated operations simpler by transforming multiplicative relationships into additive ones.
The natural logarithm introduces this by allowing us to differentiate the logarithm of a function instead of the function itself. The process involves taking the logarithm of both sides of the equation \(y = \ln(u)\), and consequently applying the properties of logarithms to simplify differentiation:

• Apply the logarithm to each part of the function, converting products and powers to sums and differences.
• Differentiate both sides of the equation.
• Solve for the derivative of the original function.

In our example, we applied this technique to the function \(y = \ln(x \sqrt{x^2 - 1})\), simplifying the differentiation process significantly. By applying logarithmic differentiation, we could transform a daunting task into a more manageable one.

Function Differentiation

Function differentiation is the backbone of calculus, interested in how a quantity changes as another quantity changes. It gives us the derivative, which tells us about the function's rate of change at any given point.
When tackling function differentiation, you need to remember the basic rules and tools, which help break down complex expressions into manageable parts, allowing them to be differentiated with ease. Core rules include:

• Power Rule for simple powers of \(x\).
• Sum Rule for the sum of functions.
• Product and Quotient Rule for products and quotients of functions.
• Chain Rule for composite functions.

In our exercise, we utilized a few of these rules to achieve the solution. Combining multiple tools and techniques ensures that we can approach any differentiation problem with confidence. Understanding and applying each of these rules thoroughly provides a strong foundation for dissecting and solving more challenging differential equations.