Problem 56 - Calculus - AP Edition

Question

Find \(d y / d x\). $$y=\csc ^{-1}\left(e^{x}\right)$$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = -\frac{1}{\sqrt{e^{2x} - 1}} \)

Step 1: Identify the Function's Derivative Formula

The function is in the form of an inverse cosecant, \( y = \csc^{-1}(u) \), where \( u = e^{x} \). To find \( \frac{dy}{dx} \), we first recall the derivative formula for \( \csc^{-1}(u) \): \[ \frac{d}{du}[ \csc^{-1}(u)] = -\frac{1}{|u|\sqrt{u^2 - 1}} \]

Step 2: Apply the Chain Rule

Because our function is composed of \( \csc^{-1} \) and an exponential function \( u = e^x \), we need to apply the chain rule to find \( \frac{dy}{dx} \). By the chain rule: \[ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \]

Step 3: Find \(\frac{dy}{du}\)

Substitute \( u = e^x \) into the derivative formula for \( \csc^{-1}(u) \):\[ \frac{dy}{du} = -\frac{1}{|e^x|\sqrt{e^{2x} - 1}} \] Since \( e^{x} > 0 \) for all real \( x \), \(|e^x| = e^x\). So, it simplifies to:\[ \frac{dy}{du} = -\frac{1}{e^x\sqrt{e^{2x} - 1}} \]

Step 4: Differentiate \(u = e^x\) with Respect to \(x\)

Compute \( \frac{du}{dx} \). Since \( u = e^x \), the derivative is:\[ \frac{du}{dx} = e^x \]

Step 5: Combine Results Using the Chain Rule

Now combine the derivatives using the chain rule from Step 2:\[ \frac{dy}{dx} = \left(-\frac{1}{e^x\sqrt{e^{2x} - 1}}\right) \times e^x \]This simplifies to:\[ \frac{dy}{dx} = -\frac{1}{\sqrt{e^{2x} - 1}} \]

Step 6: Final Expression for \(\frac{dy}{dx}\)

Thus, the derivative of the original function is:\[ \frac{dy}{dx} = -\frac{1}{\sqrt{e^{2x} - 1}} \]

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions allow us to determine angles when given trigonometric ratios. They are essential in calculus, especially when solving for derivatives involving angles. Unlike regular sine, cosine, or tangent functions, inverse trigonometric functions like \( \csc^{-1} \) (inverse cosecant) allow us to solve equations that yield an angle. In the exercise, you deal with \( \csc^{-1} \), which requires understanding of how its derivative is derived.

• For \( \csc^{-1}(u) \), the derivative is \( \frac{d}{du} [\csc^{-1}(u)] = -\frac{1}{|u|\sqrt{u^2 - 1}} \).
• This provides the rate of change of the inverse cosecant function with respect to \( u \).

This derivative is particularly important because it tells us how \( y = \csc^{-1}(u) \) changes as \( u \) changes, offering insights into the function's behavior. Mastering the derivatives of inverse trigonometric functions is key for tackling complex calculus problems involving trigonometric identities.

Chain Rule

The chain rule is a pivotal concept in calculus when dealing with composite functions—functions nested within each other. When differentiating a composite function, such as \( y = \csc^{-1}(e^x) \), the chain rule is indispensable. This rule states that to differentiate \( f(g(x)) \), you multiply the derivative of \( f \) with respect to \( g \) by the derivative of \( g \) with respect to \( x \).

• Identify the outer function \( f(u) = \csc^{-1}(u) \) and inner function \( g(x) = e^x \).
• Differentiate each: use the derivative of \( \csc^{-1}(u) \) for \( \frac{dy}{du} \) and differentiate \( e^x \) to find \( \frac{du}{dx} = e^x \).

When you multiply these derivatives together, it helps to capture the entire rate of change of the composite function regarding its original variable, \( x \). The chain rule ensures you adjust for how the inner function's change affects the outer function's change.

Exponential Functions

Exponential functions, such as \( e^x \), are fundamental in mathematics, often marked by their rapid growth. They differ from polynomial functions by maintaining a consistent rate of change proportional to their current value. Calculating derivatives for exponential functions is straightforward, particularly for the natural exponential function \( e^x \) because its derivative is also \( e^x \).

• Exponential functions have the form \( a^x \), with \( a \) being a constant base; for the natural base \( e \), \( e^x \) changes as a ratio of itself.
• For \( y = e^x \), \( \frac{dy}{dx} = e^x \).

This derivative property is crucial when it appears as a part of a more complex function, like in the exercise \( y = \csc^{-1}(e^x) \). Here, understanding exponential growth helps in applying the chain rule, as it retains its form, making calculations simpler and less error-prone. Mastery of exponential functions and their behavior is crucial when tackling calculus problems involving derivatives and rates of change.