Problem 51 - Calculus Early Transcendentals

Question

Evaluate the following derivatives. \(f(x)=x \sinh ^{-1} x-\sqrt{x^{2}+1}\)

Step-by-Step Solution

Answer

Answer: The derivative of the function is given by \(f'(x) = \sinh^{-1}x\).

Step 1: Derivative of \(x\sinh^{-1}x\) using the product rule

Recall that the product rule states that if \(u(x) = f(x)g(x)\), then the derivative of \(u(x)\) with respect to x is given by: \(u'(x) = f'(x)g(x) + f(x)g'(x)\) In this case, let \(f(x) = x\) and \(g(x) = \sinh^{-1}x\). The derivative of \(f(x)\) is \(f'(x) = 1\), while the derivative of \(g(x)\), \(\sinh^{-1}x\), is: \(g'(x) = \frac{1}{\sqrt{1+x^2}}\) Applying the product rule: \(x'(\sinh^{-1}x) = 1\sinh^{-1}x + x\frac{1}{\sqrt{1+x^2}}\)

Step 2: Derivative of \(\sqrt{x^2 + 1}\) using the chain rule

Recall the chain rule states that if \(h(x) = u(v(x))\), then the derivative of \(h(x)\) with respect to x is given by: \(h'(x) = u'(v(x))v'(x)\) In this case, let \(u(v) = \sqrt{v}\) and \(v(x) = x^2 + 1\). The derivative of \(u(v)\) is \(u'(v) = \frac{1}{2\sqrt{v}}\), while the derivative of \(v(x)\) is \(v'(x) = 2x\). Applying the chain rule: \((\sqrt{x^2 + 1})'(x) = \frac{1}{2\sqrt{x^2 + 1}}(2x) = \frac{x}{\sqrt{x^2 + 1}}\)

Step 3: Combining the two derivatives

Now let's combine the results from step 1 and step 2 to find the derivative of the function \(f(x)\): \(f'(x) = (x\sinh^{-1}x)' - (\sqrt{x^2 + 1})'\) \(= \left(1\sinh^{-1}x + x\frac{1}{\sqrt{1+x^2}}\right) - \frac{x}{\sqrt{x^2 + 1}}\)

Step 4: Simplify the derivative

Finally, simplify the derivative: \(f'(x) = \sinh^{-1}x + x\frac{1}{\sqrt{1+x^2}} - \frac{x}{\sqrt{x^2 + 1}}\) \(f'(x) = \sinh^{-1}x\)

Key Concepts

Understanding the Product Rule

The product rule is a core concept in calculus that is essential when taking the derivative of a product of two functions. Suppose you have a function that is composed of two sub-functions, say, \( u(x) = f(x)g(x) \). To differentiate this type of function, you cannot simply take the derivative of each part separately. That's where the product rule comes in handy.

The product rule states:

• If \( u(x) = f(x)g(x) \), then the derivative of \( u(x) \) is \( u'(x) = f'(x)g(x) + f(x)g'(x) \).

This rule helps you find the total derivative by considering both parts of the product: the derivative of \( f(x) \) times \( g(x) \), plus \( f(x) \) times the derivative of \( g(x) \).

For example, in the exercise given, \( f(x) = x \) and \( g(x) = \sinh^{-1}x \). Using the product rule, the derivative becomes:

• \( 1\cdot \sinh^{-1}x + x\cdot \frac{1}{\sqrt{1+x^2}} \).

This method ensures you correctly account for both components of the product in the differentiation process.

Delving into the Chain Rule

The chain rule is another fundamental tool in calculus used for finding the derivative of a composite function. When you have a function composed of another function, this is your go-to technique.

Let's say you have a function \( h(x) = u(v(x)) \), where \( u \) is a function of \( v \) and \( v \) is a function of \( x \). The chain rule will help you differentiate \( h(x) \) by multiplying the derivative of the outer function evaluated at the inner function by the derivative of the inner function.

The chain rule can be expressed as:

• \( h'(x) = u'(v(x)) \cdot v'(x) \).

This equation tells us that the derivative of \( h(x) \) involves evaluating the derivative of the outer function at the inner function, and then multiplying by the derivative of the inner function.

Taking an example from the exercise, if you have \( \sqrt{x^2 + 1} \), you apply the chain rule as follows:

• Set \( u(v) = \sqrt{v} \) and \( v(x) = x^2 + 1 \).
• Find \( u'(v) = \frac{1}{2\sqrt{v}} \) and \( v'(x) = 2x \).
• Using the chain rule: \( \frac{1}{2\sqrt{x^2 + 1}} \times 2x = \frac{x}{\sqrt{x^2 + 1}} \).

This rule clarifies how each layer of the function is differentiated in sequence by applying the derivatives step-by-step.

Exploring Inverse Hyperbolic Functions

Inverse hyperbolic functions are extensions of hyperbolic functions, similar to how inverse trigonometric functions relate to trigonometric functions. They are vital in various mathematical models and calculations.

One inverse hyperbolic function is \( \sinh^{-1}(x) \), also known as the inverse hyperbolic sine. If \( y = \sinh^{-1}(x) \), then \( x = \sinh(y) \). This relationship is solved by expressing \( y \) in terms of \( x \).

To derive \( \sinh^{-1}(x) \), it's crucial to understand its derivative:

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

This formula originates from differentiating the implicit relationship and solving for the derivative. In the original exercise, this derivative played a part in applying the product rule with the function \( x\sinh^{-1}(x) \).

The use of inverse hyperbolic functions allows mathematicians and engineers to handle equations involving hyperbolic constants more efficiently, simplifying calculations in fields such as physics and hyperbolic geometry.