Problem 45 - Calculus and its applications

Question

Differentiate. $$ y=\sqrt{e^{x}-1} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{e^x}{2\sqrt{e^x - 1}} \).

Step 1: Rewrite the Function

First, express the original function in terms of an exponential and power function. We have:\[y = (e^x - 1)^{1/2}\]This makes it easier to apply the chain rule for differentiation.

Step 2: Apply the Chain Rule

Differentiate the outer function, which is \((u)^{1/2}\), where \(u = e^x - 1\). The derivative is:\[\frac{d}{du}(u^{1/2}) = \frac{1}{2}u^{-1/2}\]This means:\[\frac{d}{dx}(e^x - 1)^{1/2} = \frac{1}{2}(e^x - 1)^{-1/2} \times \frac{d}{dx}(e^x - 1)\]

Step 3: Differentiate the Inner Function

Now, find the derivative of \(u = e^x - 1\):\[\frac{d}{dx}(e^x - 1) = \frac{d}{dx}(e^x) - \frac{d}{dx}(1) = e^x - 0 = e^x\]

Step 4: Combine Using the Chain Rule

Substitute the derivative of the inner function into the expression found in Step 2:\[\frac{d}{dx}(e^x - 1)^{1/2} = \frac{1}{2}(e^x - 1)^{-1/2} \times e^x\]This simplifies to:\[\frac{e^x}{2\sqrt{e^x - 1}}\]

Key Concepts

Chain Rule

When differentiating composite functions, the chain rule is an essential tool in calculus. It helps us differentiate functions that are nested within other functions.

The chain rule states that if you have a composite function, say \( f(g(x)) \), then the derivative \( f'(x) \) is the derivative of \( f \) with respect to \( g \) multiplied by the derivative of \( g \) with respect to \( x \). Mathematically, this is:

• \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \)

In our exercise, the outer function is \((u)^{1/2}\) where \(u = e^x - 1\). We differentiate \((u)^{1/2}\), using the power rule, which comes in combination with the chain rule here.

We then multiply it by the derivative of the inner function \(e^x - 1\), which is just the derivative of \(e^x\), making the process straightforward.

Exponential Function Derivative

Understanding the derivative of exponential functions is crucial, as these functions appear frequently in calculus. The exponential function \( e^x \), where \( e \) is Euler's number, has a unique property: its derivative is the same as the function itself.

• The derivative of \( e^x \) is simply \( e^x \).

In our given example, the function is expressed in terms of a square root of a difference involving an exponential function.

For the inner function \( e^x - 1 \), it helps to know that subtracting a constant does not affect the derivative, so:

• \( \frac{d}{dx}(e^x - 1) = \frac{d}{dx}(e^x) - \frac{d}{dx}(1) = e^x - 0 = e^x \)

This understanding aids in applying the derivative correctly when using the chain rule.

Power Rule in Calculus

The power rule in calculus is one of the most fundamental differentiation rules and is often combined with other rules, such as the chain rule, to differentiate more complex functions. The power rule states:

• If \( y = x^n \), then \( \frac{dy}{dx} = nx^{n-1} \).

In our exercise, after rewriting the original problem using exponent notation, we use the power rule to solve for the derivative of \((u)^{1/2}\), where \( u = e^x - 1 \). This gives:

• \( \frac{d}{du}(u^{1/2}) = \frac{1}{2}u^{-1/2} \)

Once the power rule is applied, the result is then multiplied by the derivative of the inner function according to the chain rule.
This demonstrates how powerful and flexible the power rule is when dealing with functions raised to any power, even fractions like \( \frac{1}{2} \), enabling us to simplify otherwise complex differentiation tasks.