Problem 410 - Calculus Volume 1

Question

For the following exercises, find the derivatives for the functions. $$e^{\sinh ^{-1}(x)}$$

Step-by-Step Solution

Answer

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

Step 1: Understand the Function

The function given is an exponential function where the exponent is the inverse hyperbolic sine of \(x\). It is written as \(y = e^{\sinh^{-1}(x)}\). To find the derivative, we'll use the chain rule.

Step 2: Recall the Chain Rule

The chain rule states that if you have a composite function \(f(g(x))\), its derivative is \(f'(g(x)) \cdot g'(x)\). Here, \(f(u) = e^u\) and \(u = \sinh^{-1}(x)\). So, the derivative will be \(e^{\sinh^{-1}(x)}\) multiplied by the derivative of \(\sinh^{-1}(x)\).

Step 3: Find the Derivative of the Inner Function

The derivative of \(\sinh^{-1}(x)\) with respect to \(x\) is \(\frac{1}{\sqrt{x^2 + 1}}\). This is a standard derivative for inverse hyperbolic sine, which can be derived from its definition or known from derivative tables.

Step 4: Apply the Chain Rule

Applying the chain rule, we multiply the derivative of the outer function \(e^u\) by the derivative of the inner function: \(\frac{d}{dx} (e^{\sinh^{-1}(x)}) = e^{\sinh^{-1}(x)} \cdot \frac{1}{\sqrt{x^2 + 1}}\).

Step 5: Simplify the Derivative

Since there are no further simplifications needed, the derivative of the given function remains \(e^{\sinh^{-1}(x)} \cdot \frac{1}{\sqrt{x^2 + 1}}\).

Key Concepts

Chain Rule

The chain rule is essential when differentiating composite functions. Composite functions are those where one function is nested inside another. The rule simplifies finding derivatives by linking the derivative of the outer function to that of the inner function.
The formula for the chain rule is:

• If a function is written as \( f(g(x)) \), its derivative is \( f'(g(x)) \cdot g'(x) \).

In the context of the exercise, we have the function \( y = e^{\sinh^{-1}(x)} \). Here, the outer function is \( f(u) = e^u \) and the inner function is \( u = \sinh^{-1}(x) \). So, when applying the chain rule, we find the derivative of the outer function while keeping the inner function constant, and multiply it by the derivative of the inner function.
Using the chain rule allows us to tackle the differentiation of more complex structures by breaking them down into simpler parts.

Inverse Hyperbolic Functions

Inverse hyperbolic functions are counterparts to the regular hyperbolic functions, analogous to how inverse trigonometric functions relate to their standard counterparts. The inverse hyperbolic sine, \( \sinh^{-1}(x) \), is particularly important in calculus for simplifying integrals and derivatives.

• The derivative of \( \sinh^{-1}(x) \) with respect to \( x \) is \( \frac{1}{\sqrt{x^2 + 1}} \).

This derivative can be derived from the definition of \( \sinh(x) \) or directly from tables of standard derivatives. The function \( \sinh^{-1}(x) \), also known as \( \text{arsinh}(x) \), helps when calculating the slope of curves involving hyperbolic relationships. In the exercise, recognizing \( \sinh^{-1}(x) \) allows us to make use of its standard derivative rule, simplifying the overall differentiation process.

Exponential Functions

Exponential functions are fundamental in calculus and appear in many areas of mathematics and natural sciences. An exponential function is one where a constant base, usually \( e \), is raised to a variable exponent.

• The expression \( e^{\sinh^{-1}(x)} \) involves raising \( e \) to the power of an inverse hyperbolic function.
• The derivative of an exponential function \( e^x \) is simply \( e^x \).

This unique property makes them easy to differentiate. In the problem at hand, the function \( e^{\sinh^{-1}(x)} \) is differentiated by keeping the base \( e \), but we need to multiply by the derivative of the exponent \( \sinh^{-1}(x) \). Exponential functions often model growth and decay processes but, in this case, they combine complex elements and demonstrate the power of the chain rule in finding derivatives.