Question

Find the derivatives of the following functions. $$f(v)=\sinh ^{-1} v^{2}$$

Short answer

Question: Find the derivative of the function \(f(v) = \sinh^{-1}(v^2)\) with respect to \(v\). Answer: The derivative of the given function \(f(v) = \sinh^{-1}(v^2)\) with respect to \(v\) is \(\frac{df}{dv} = \frac{2v}{\sqrt{1 + v^4}}\).

Step-by-step solution

Identify the functions g(v) and f(u)

First, we can write the given function as a composition of two functions: \(g(v) =v^2\) and \(f(u) =\sinh ^{-1}(u)\), where \(u = g(v)\).

Differentiate g(v) and f(u) with respect to their variables

We will now differentiate the functions \(g(v)\) and \(f(u)\) with respect to their respective variables \(v\) and \(u\): 1. Differentiate \(g(v)\) with respect to \(v\): $$\frac{dg}{dv} = \frac{d(v^2)}{dv} = 2v$$ 2. Differentiate \(f(u)\) with respect to \(u\): $$\frac{df}{du} = \frac{d(\sinh^{-1}(u))}{du} = \frac{1}{\sqrt{1 + u^2}}$$

Apply the chain rule

Now, we can apply the chain rule to find the derivative of \(f(v)\) with respect to \(v\). We have \(\frac{df}{dv} = \frac{df}{du} \cdot \frac{du}{dv}\), where \(u = g(v)\). Substitute \(g(v)\), \(\frac{dg}{dv}\) and \(\frac{df}{du}\) into the chain rule formula: $$\frac{df}{dv} = \left(\frac{1}{\sqrt{1 + u^2}}\right) \cdot \left( 2v \right) = \frac{2v}{\sqrt{1 + u^2}}$$ Since \(u = g(v) = v^2\), we can substitute \(u\) back: $$\frac{df}{dv} = \frac{2v}{\sqrt{1 + v^4}}$$

Final Answer

The derivative of the given function \(f(v) = \sinh^{-1}(v^2)\) with respect to \(v\) is: $$\frac{df}{dv} = \frac{2v}{\sqrt{1 + v^4}}$$

Key concepts

Chain Rule

The chain rule is a fundamental concept in calculus used for finding the derivative of a composition of two or more functions. Imagine you have a function nested inside another function, such as \(f(g(x))\). To find the derivative of this composite function, the chain rule helps you differentiate them step by step.

Here's a simple way to think about the chain rule:

• Identify the inner function and the outer function.
• Differentiate the outer function with respect to the inner function.
• Differentiate the inner function with respect to its own variable.
• Multiply these derivatives together.

In the given exercise, the outer function is \(\sinh^{-1}(u)\) and the inner function is \(v^2\). By applying the chain rule, we first differentiate the outer function \(f(u)\) with respect to \(u\), and then we differentiate the inner function \(g(v)\) with respect to \(v\). Finally, we multiply the two results to obtain the derivative of \(f(v)\).

This approach allows us to handle complex, layered functions simply by breaking them down step by step.

Inverse Hyperbolic Functions

Inverse hyperbolic functions, like \(\sinh^{-1}(x)\), are related to the standard hyperbolic functions, but instead of mapping real numbers to hyperbolic coordinates, they do the reverse. They are essential in calculus for solving equations involving hyperbolic functions.

Some of the common inverse hyperbolic functions include:

• \(\sinh^{-1}(x)\) - the inverse hyperbolic sine function.
• \(\cosh^{-1}(x)\) - the inverse hyperbolic cosine function.
• \(\tanh^{-1}(x)\) - the inverse hyperbolic tangent function.

When differentiating inverse hyperbolic functions, we use specific formulas. For \(\sinh^{-1}(x)\), the derivative is \(\frac{1}{\sqrt{1+x^2}}\). By applying these formulas, we can find the rate at which the function changes as its variable changes. This makes inverse hyperbolic functions a powerful tool for analyzing hyperbolic relationships in mathematical problems.

Composition of Functions

Composition of functions involves combining two functions to form a new one. The composition is achieved by plugging one function into another, such as \(f(g(x))\), where \(g(x)\) is evaluated first and its result is used as the input for \(f(x)\).

Here's how to think of compositions:

• Consider it as a process where the output of one function becomes the input for the next.
• This creates a layering of functions, allowing complex relations to be simplified step by step.

In the exercise, we took the composition \(\sinh^{-1}(v^2)\). First, \(v^2\) is computed and becomes the input for \(\sinh^{-1}\). Understanding this layering helps in differentiating the function using methods like the chain rule, as it clearly defines the step where each function is applied.

Compositions are often found in real-world applications where processes depend on each other, demonstrating how one variable influences multiple results through a sequence of operations.