Problem 36 - Single Variable Calculus

Question

\(y=e^{\sinh 2 t}\)

Step-by-Step Solution

Answer

The derivative of \(y = e^{\sinh (2t)}\) is \(y' = 2e^{\sinh (2t)} \cdot \cosh(2t)\).

Step 1: Understanding the Functions

Before starting to compute the derivative, it's important to understand the functions involved. \( e^x \) is the exponential function where \( e \) is the base of the natural logarithm. \(\sinh(x)\) is the hyperbolic sine function, defined by the formula \(\sinh(x) = \frac{e^x - e^{-x}}{2}\).

Step 2: Apply the Chain Rule

We will use the chain rule to obtain the derivative. The chain rule states that the derivative of the composition of two functions is the derivative of the first function evaluated at the inside function times the derivative of the inside function. If we have a function like \(y = f(g(x))\), then the derivative \(y'\) is given by \(f'(g(x)) * g'(x)\). Here, \(y = e^{\sinh(2t)}\), which is in the form \(f(g(t))\), with \(f(u) = e^u\) and \(g(t) = \sinh(2t)\).

Step 3: The Derivative of the Inner Function

First we calculate the derivative of the inner function, \(g(t) = \sinh(2t)\). The derivative of \( \sinh(x) \) is \( \cosh(x) \), and for any constant \( a \), the derivative of \( a \cdot t \) is \( a \). Therefore, the derivative of \( \sinh(2t) \), is \( 2 \cdot \cosh(2t)\).

Step 4: The Derivative of the Outer Function

Next, we calculate the derivative of the outer function, \(f(u) = e^u\), with respect to \(u\). The derivative of the exponential function \( e^u \) is also \( e^u \), so the derivative of \(f\) at \(u = \sinh(2t)\) is \(e^{\sinh (2t)}\).

Step 5: Apply the Chain Rule

Now we can calculate the derivative of \(y = e^{\sinh (2t)}\) by multiplying the derivative of the outer function by the derivative of the inner function, per the chain rule. This gives us \( y' = e^{\sinh (2t)} \cdot 2cosh(2t) \).

Step 6: Simplify the Result

The final derivative of the function is therefore \(y' = 2e^{\sinh (2t)} \cdot \cosh(2t) \).

Key Concepts

Chain Rule

When you come across a problem involving the derivative of a composite function, the chain rule is your go-to tool. The chain rule helps us find the derivative of functions nested within each other. Think of it as peeling away layers of an onion, where each layer is a function.
For instance, if we have a function like

• \( y = e^{\sinh(2t)} \;\)

Here you see two layers:

• **The outer layer**: Exponential function \( e^u \);
• **The inner layer**: Hyperbolic sine function \( \sinh(2t) \);

To find the derivative of \( y \), we take the derivative of the outer function while keeping the inner function constant. Then, multiply it by the derivative of the inner function. Each derivative gives different clues about how changes in the input affect the entire structure of the function.
This allows us to approach each part systematically, thereby making the process more manageable. By applying the chain rule, you will ensure all necessary changes between the layers are accounted for.

Hyperbolic Functions

Hyperbolic functions provide fascinating insights into calculus, often used to model real-world dynamics like waves and oscillations. The function used in our exercise, \(\sinh(x)\), is one such hyperbolic function, defined as:

• \( \sinh(x) = \frac{e^x - e^{-x}}{2} \);

The hyperbolic sine function, \(\sinh(x)\), forms the crux of many calculus examples because its derivatives lead to other hyperbolic functions, creating an interconnected set of formulas.
When taking derivatives of these functions, it's helpful to know:

• \(\text{Derivative of } \sinh(x)\text{ is } \cosh(x) \);
• \(\text{Derivative of } \cosh(x)\text{ is } \sinh(x) \);

These relationships make solving calculus problems more intuitive since similar rules apply across the family of hyperbolic functions. Each function behaves predictably, much like their trigonometric counterparts, giving calculus its robust consistency.

Derivative Calculation

In calculus, calculating derivatives is a way to ascertain how a function changes as its input changes. For our particular exercise, we've been working with the function \( y = e^{\sinh(2t)} \). Let's break it down step by simple step.
Firstly, identify the functions involved. We already know there's an exponential function with the hyperbolic sine of \( 2t \) nested inside. Knowing the basic derivatives helps greatly:

• \(\text{Derivative of } e^u \text{ is } e^u \);
• \(\text{derivative of } \sinh(x)\text{ is } \cosh(x) \);
• \(\text{Constant multiplier rule: if } g(t) = 2t, g'(t) = 2 \);

Using chain rule, the derivative of \( y \) is the derivative of the outer function, \( e^{\sinh(2t)} \), which is \( e^{\sinh(2t)} \). Multiply it by the derivative of the inner function \( \sinh(2t) \)—that's \( 2\cosh(2t) \). Combining these through multiplication, the derivative is:

• \( y' = 2e^{\sinh(2t)} \cdot \cosh(2t) \)

This calculation shows how each component contributes to the rate of change of \( y \). By understanding the process, you'll be better prepared for tackling even more complex derivatives in the future.