Question

Find the derivatives of the given functions and graph along with the function to ensure your answer is correct.\([\mathrm{T}] \sinh (\ln (x))\)

Short answer

Derivative: \(\frac{\cosh(\ln(x))}{x}\)

Step-by-step solution

Understand the Problem

We need to find the derivative of the function \(\sinh(\ln(x))\). \(\sinh(u)\) is the hyperbolic sine function defined as \(\sinh(u) = \frac{e^u - e^{-u}}{2}\). Our goal is to differentiate \(\sinh(\ln(x))\) with respect to \(x\).

Use the Chain Rule

We will use the chain rule to differentiate \(\sinh(\ln(x))\). The chain rule states that if we have a composite function \(f(g(x))\), then the derivative is \(f'(g(x)) \cdot g'(x)\).

Differentiate \(\sinh(u)\) with respect to \(u\)

The derivative of \(\sinh(u)\) with respect to \(u\) is \(\cosh(u)\). So, for our function, \(\sinh(\ln(x))\), differentiate with respect to \(\ln(x)\) yields \(\cosh(\ln(x))\).

Differentiate \(\ln(x)\) with respect to \(x\)

The derivative of \(\ln(x)\) with respect to \(x\) is \(\frac{1}{x}\).

Combine the results using the Chain Rule

Combine the derivatives from Steps 3 and 4 using the chain rule: \[ \frac{d}{dx} \sinh(\ln(x)) = \cosh(\ln(x)) \cdot \frac{1}{x} = \frac{\cosh(\ln(x))}{x} \].

Graph the Function and Its Derivative

To ensure accuracy, graph both \(f(x) = \sinh(\ln(x))\) and its derivative \(f'(x) = \frac{\cosh(\ln(x))}{x}\). Observe the relationship; the derivative should provide the slope of the tangent lines to the original function at each point.

Key concepts

Chain Rule

In calculus, the chain rule is a fundamental tool for finding the derivative of a composite function. A composite function is basically a function within another function. The chain rule provides a way to differentiate such functions by breaking them down into simpler parts.
To apply the chain rule, consider a function of the form \(f(g(x))\). Here, \(g(x)\) is nested inside \(f\). According to the chain rule:

• First, differentiate the outer function \(f\) with respect to the inside function \(g(x)\), giving \(f'(g(x))\).
• Next, differentiate the inside function \(g(x)\) with respect to \(x\), giving \(g'(x)\).
• Finally, multiply the two derivatives from the steps above to get \(f'(g(x)) \cdot g'(x)\).

For example, in our exercise, to find the derivative of \(\sinh(\ln(x))\), we consider \(\sinh(u)\) as the outer function and \(\ln(x)\) as the inner function. By finding these derivatives and multiplying them, we successfully apply the chain rule. This step allows us to move from complex to simple, piece by piece, which is what makes calculus so powerful.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but for the hyperbola, much like trigonometric functions relate to the circle. Two of the most important hyperbolic functions are the hyperbolic sine \(\sinh\) and the hyperbolic cosine \(\cosh\).
The hyperbolic sine function is defined as:
\[ \sinh(u) = \frac{e^u - e^{-u}}{2} \]
where \(e\) is the base of the natural logarithm. When you differentiate \(\sinh(u)\) with respect to \(u\), you get the hyperbolic cosine, \(\cosh(u)\):
\[ \cosh(u) = \frac{e^u + e^{-u}}{2} \]
These functions have several interesting properties:

• They have exponential forms, relating them to the exponential function \(e^x\).
• They exhibit periodic oscillations, similar to trigonometric functions, but over real numbers.

In our problem, \(\sinh(\ln(x))\) involves both a hyperbolic function and a logarithm. This mixture demonstrates the versatility of calculus, highlighting how different mathematical ideas intertwine to form complex and useful functions.

Graphing Functions

Graphing functions helps us visually understand their behavior and characteristics. It allows us to see the relationship between a function and its derivative, which can be invaluable for checking solutions.

When we graph a function like \(f(x) = \sinh(\ln(x))\) and its derivative \(f'(x) = \frac{\cosh(\ln(x))}{x}\), we can observe several things:

• The derivative graph demonstrates the slope of tangent lines at different points on the original function's graph.
• Peaks and valleys on the \(f(x)\) graph suggest where \(f'(x)\) crosses the x-axis, indicating zero slopes.
• Understanding these graphs helps in identifying behavior such as increasing/decreasing trends and concavity.

Thus, graphing not only confirms our algebraic differentiation but also provides deeper insights into how the function behaves across its domain. Visualization is a key aspect of understanding more elaborate functions and their attributes in calculus.