Question

Evaluate the following derivatives. \(f(v)=\sinh ^{-1} v^{2}\)

Short answer

Answer: The derivative of the function is \(f'(v) = \frac{2v}{\sqrt{1 + v^4}}\).

Step-by-step solution

Find the derivative of g(u)

The derivative of \(\sinh^{-1}(u)\) with respect to \(u\) is given by: \(g'(u) = \frac{1}{\sqrt{1 + u^2}}\)

Find the derivative of h(v)

The derivative of \(v^2\) with respect to \(v\) is given by: \(h'(v) = 2v\)

Apply the chain rule

According to the chain rule, the derivative of \(f(v)\) is given by: \(f'(v) = g'(h(v)) \cdot h'(v)\) Substitute \(g'(u)\) and \(h'(v)\) values found in the previous steps and replace \(h(v)\) with \(v^2\): \(f'(v) = \frac{1}{\sqrt{1 + (v^2)^2}} \cdot 2v\)

Simplify the expression

Simplify the expression for \(f'(v)\): \(f'(v) = \frac{2v}{\sqrt{1 + v^4}}\) Thus, the derivative of \(f(v) = \sinh^{-1}(v^2)\) with respect to \(v\) is \(f'(v) = \frac{2v}{\sqrt{1 + v^4}}\).

Key concepts

Derivatives

In calculus, derivatives represent the rate of change of a function concerning its variable. Simply put, a derivative tells us how a function's output value changes when the input changes. Think of it as asking, "How fast is something changing?" This is a fundamental concept in calculus, used to understand behaviors of functions and solve complex mathematical problems. Einstein famously used calculus to develop the Theory of Relativity, which includes computing derivatives of various functions! In our exercise, we are dealing with the inverse hyperbolic sine function, denoted as \( \sinh^{-1}(u) \). The derivative here specifically is \( g'(u) = \frac{1}{\sqrt{1 + u^2}} \), which tells us how the inverse hyperbolic sine function changes as \( u \) changes.

Chain Rule

The chain rule is a crucial technique for finding derivatives when dealing with compositions of functions. Many real-world problems use functions that consist of multiple internal functions. The chain rule lets us divide the problem into small, manageable parts. Essentially, the chain rule states: if you have two functions \( g(u) \) and \( h(v) \), the derivative of their composition \( f(v) = g(h(v)) \) is given by: \[ f'(v) = g'(h(v)) \cdot h'(v) \] This rule allows us to differentiate complex functions methodically. In our specific case, we apply the chain rule to differentiate a composition of the inverse hyperbolic sine function with the square function, using the derivatives we calculated for both.

Function Composition

Function composition involves nesting one function inside another. This concept is useful when combining different processes to be represented as a single operation. It answers the question: "What happens if I apply one function to a number, then immediately apply another?"For instance, let two functions be \( g(u) = \sinh^{-1}(u) \) and \( h(v) = v^2 \). Then, their composition is \( f(v) = g(h(v)) \) or \( f(v) = \sinh^{-1}(v^2) \). By understanding function composition, we can solve intricate mathematical problems by breaking them down into simpler steps. It allows us to approach complex derivative problems in simpler terms, using known derivative results for composited functions such as \( g \) and \( h \) separately.