Problem 23 - Essential Calculus

Question

Find the derivative of the function. \(f(t)=\arctan (\sinh t)\)

Step-by-Step Solution

Answer

The derivative of the function \(f(t)=\arctan (\sinh t)\) is \(f'(t) = \frac {\cosh t} {1 + (\sinh t)^2}\).

Step 1: Find the Derivative of Outer Function

Consider \(f(t) = \arctan (u)\) where \(u = \sinh t\). The derivative of the outer function in terms of \(u\) follows as \(f'(t) = \frac {1} {1+ u^2}\). Substitute \(u\) back as \(\sinh t\) to get \(f '(t) = \frac {1} {1 + (\sinh t)^2}\).

Step 2: Find the Derivative of Inner Function

Now, find the derivative of the inner function \(u = \sinh t\). This differentiates to \(u'= \cosh t\).

Step 3: Apply the Chain Rule

Use the chain rule, which states that \( \frac {d} {dt}(\arctan (\sinh t)) = f '(t) \cdot u' \). So the derivative of \( \arctan (\sinh t) \) is \( \frac {1} {1 + (\sinh t)^2} \cdot \cosh t = \frac {\cosh t} {1 + (\sinh t)^2}\).

Key Concepts

Chain Rule in Calculus

The chain rule is a fundamental technique in calculus used for finding the derivative of composite functions. In simpler terms, it's a way to differentiate functions that are nestled inside each other. For example, if a function 'h' consists of 'f(g(t))', then the derivative of 'h' with respect to 't' is the derivative of 'f' with respect to 'g(t)' times the derivative of 'g' with respect to 't', expressed as
\(h'(t) = f'(g(t)) \cdot g'(t)\).
This formula helps break down complex differentiation problems into more manageable parts. The first step typically involves taking the derivative of the outer function, as if the inner function were a simple variable. Afterward, the derivative of the inner function is found and multiplied by the derivative of the outer function.

• Identify the outer and inner functions
• Compute the derivative of the outer function (leave the inner function unchanged)
• Find the derivative of the inner function separately
• Multiply the derivatives from the previous steps

In the example of \(f(t)=\arctan (\sinh t)\), the outer function is \(\arctan(u)\) and the inner function is \(u=\sinh t\). According to the chain rule, we'll first differentiate \(\arctan(u)\) with respect to 'u', and then multiply it by the derivative of \(\sinh t\) with respect to 't'.

Hyperbolic Functions

Hyperbolic functions can be considered as analogs of the trigonometric functions but for a hyperbola instead of a circle. The two most basic hyperbolic functions are the hyperbolic sine and cosine, denoted by \(\sinh t\) and \(\cosh t\), respectively. Similar to trigonometric functions, hyperbolic functions have several useful properties and identities that make them important in calculus.
\(\sinh t\) is defined as \(\frac{e^{t} - e^{-t}}{2}\), and \(\cosh t\) is defined as \(\frac{e^{t} + e^{-t}}{2}\). These functions are useful in various areas, including the calculation of lengths and angles in hyperbolic geometry, and in solving certain differential equations.
The derivatives of these functions are straightforward: the derivative of \(\sinh t\) with respect to 't' is \(\cosh t\), and the derivative of \(\cosh t\) with respect to 't' is \(\sinh t\).

• \(\sinh' t = \cosh t\)
• \(\cosh' t = \sinh t\)

These derivatives are applied in the chain rule when differentiating functions involving hyperbolic sine or cosine.

Implicit Differentiation

Implicit differentiation is a technique used to find the derivative of a function that is not explicitly solved for one variable in terms of another. In other words, if the function is given in an implicit form, where both 'x' and 'y' are mixed together, implicit differentiation allows us to calculate the derivative of 'y' with respect to 'x' without isolating 'y'.
This approach involves differentiating each term of the equation with respect to 'x' while treating 'y' as a function of 'x' (if applicable), and using the chain rule when necessary. Here's how to apply it:

• Differentiate both sides of the equation with respect to 'x'.
• Apply the chain rule when differentiating terms that include 'y'.
• Solve the resulting equation for \(\frac{dy}{dx}\) to find the derivative of 'y' with respect to 'x'.

Implicit differentiation is particularly helpful in problems where it's difficult or unnecessary to solve for one variable before differentiating, such as when dealing with equations of circles, ellipses, or hyperbolas. In the example \(f(t)=\arctan (\sinh t)\), the function is already given explicitly in one variable 't', so implicit differentiation isn't necessary. However, understanding this concept enriches one's differentiation toolkit when faced with a broader range of problems.