Question

Find the derivatives of the given functions and graph along with the function to ensure your answer is correct.\([T] \sinh \left(x^{2}\right)\)

Short answer

The derivative of \( \sinh(x^2) \) is \( 2x \cosh(x^2) \).

Step-by-step solution

Understand the Hyperbolic Sine Function

The function given is the hyperbolic sine of a function, specifically \( \sinh(x^2) \). The derivative of the hyperbolic sine \( \sinh(u) \) with respect to \( u \) is \( \cosh(u) \). In our case, \( u = x^2 \).

Apply the Chain Rule

To find the derivative of \( \sinh(x^2) \), we need to apply the chain rule. This rule states that if you have a composite function \( f(g(x)) \), its derivative is \( f'(g(x)) \cdot g'(x) \).

Differentiate the Outer Function

Differentiate the outer function \( \sinh(u) \) with respect to \( u \), which gives \( \cosh(u) \). So the derivative with respect to \( x \) will be \( \cosh(x^2) \cdot \frac{d}{dx}(x^2) \).

Differentiate the Inner Function

Differentiate the inner function \( x^2 \) with respect to \( x \), resulting in \( 2x \).

Combine the Derivatives

Multiply the derivative of the outer function by the derivative of the inner function: \[ \frac{d}{dx} \sinh(x^2) = \cosh(x^2) \cdot 2x \].

Verify by Graphing

To verify the result, graph both the original function \( \sinh(x^2) \) and the derivative \( 2x \cosh(x^2) \). The derivative graph should reflect the slopes of \( \sinh(x^2) \) at various points.

Key concepts

Derivative

The derivative is a fundamental concept in calculus, which measures how a function changes as its input changes. It's like calculating the slope of a curve. For any function, the derivative tells us how steep the curve is at any given point, indicating the rate of change.
In the context of our exercise, we dealt with differentiating the hyperbolic sine function, specifically a composite function \(\sinh(x^2)\). Finding its derivative involves several steps, including applying other calculus rules, such as the chain rule.
When we differentiate, we're essentially finding the function's instantaneous rate of change. For instance, when we differentiated \(\sinh(x^2)\), we were not just looking at the direct change but also accounting for how \(x^2\), as an inner component, affected the overall rate of change. Understanding derivatives is crucial for everything from physics to economics because they model real-world scenarios where things change continuously.

Hyperbolic Functions

Hyperbolic functions are analogous to the trigonometric functions but for hyperbolas instead of circles. While you might be more familiar with sine, cosine, or tangent, hyperbolic functions provide useful properties and are vital in many areas, including calculus and hyperbolic geometry.
The hyperbolic sine function, \( \sinh(x) \), is denoted as \( \frac{e^x - e^{-x}}{2} \). It's important to remember this definition, as it helps when differentiating or integrating hyperbolic functions. In our exercise, we focused on \( \sinh(x^2) \), which is a transformation of the hyperbolic sine by squaring its argument.
Knowing the derivatives of hyperbolic functions is essential because these functions often appear in various scientific models, including those describing the shape of a hanging cable or the distribution of light intensity in optics. By applying calculus, especially the chain rule, we can uncover how these functions change and influence the systems they model.

Chain Rule

The chain rule is an essential tool in calculus for differentiating composite functions. It allows us to find the derivative of a function nested inside another function, like \(f(g(x))\). This scenario occurs often in calculus problems.
The rule states that the derivative of \(f(g(x))\) is \(f'(g(x)) \cdot g'(x)\). Understanding this method is crucial for handling functions like \( \sinh(x^2) \), where two functions combine into one. Here, \( f(u) = \sinh(u) \) and \( g(x) = x^2 \), allowing us to apply the chain rule seamlessly.
Using the chain rule effectively means identifying the outer and inner functions accurately. Once we split the composite function, we tackle the derivative of the outer function, \(f(u)\), first, and then multiply it with the derivative of the inner function, \(g(x)\). The chain rule simplifies the process and unveils the nature of changes within complex functions, proving invaluable in science and engineering problems.