Problem 59 - Essential Calculus

Question

Find the derivative of the function. \(y=2 x \sinh ^{-1}(2 x)-\sqrt{1+4 x^{2}}\)

Step-by-Step Solution

Answer

The derivative \(y'\) is \(2*sinh^{-1}(2x) + 2x*\frac{1}{\sqrt{1+(2x)^2}}*2 - \frac{1}{2}*(1+4x^2)^{-\frac{1}{2}} * 8x\).

Step 1: Find the Derivative of the First Part

The first part is \(2x sinh^{-1}(2x)\) which can be seen as a product between two functions. Therefore, the product rule of differentiation will be applied. The product rule is \(u'v + v'u\). Let \(u=2x\) and \(v=\sinh^{-1}(2x)\). The derivative \(u'\) is 2. The derivative \(v'\) of \(v=\sinh^{-1}(2x)\) is \(\frac{1}{\sqrt{1+(2x)^2}}*2\). Thus, the derivative of the first part is \(2*sinh^{-1}(2x) + 2x*\frac{1}{\sqrt{1+(2x)^2}}*2\)

Step 2: Find the Derivative of the Second Part

The second term is \(-\sqrt{1+4x^2}\). The derivative of the square root function generally is \(-\frac{1}{2}*(1+4x^2)^{-\frac{1}{2}} * 8x\). This is by applying the chain rule for derivatives

Step 3: Combine These Two Derivatives

The overall derivative is obtained by combining these two derivatives obtained from step 1 and step 2. Therefore, the derivative \(y'\) is \(2*sinh^{-1}(2x) + 2x*\frac{1}{\sqrt{1+(2x)^2}}*2 - \frac{1}{2}*(1+4x^2)^{-\frac{1}{2}} * 8x\). This can be simplified further.

Key Concepts

Product Rule of Differentiation

One of the fundamental rules in calculus is the product rule, which enables us to find the derivative of two multiplied functions. This rule is essential when we encounter an expression where two variable expressions are multiplied together, as in the case of the function given in the exercise: \( y=2x\sinh^{-1}(2x) \).

According to the product rule, if we have two functions, \(u(x)\) and \(v(x)\), their derivative \(w'(x)\) when multiplied \(w(x)=u(x)v(x)\) is given by: \[ w'(x)=u'(x)v(x)+v'(x)u(x) \].

In the context of the exercise, we apply this rule by identifying \(u=2x\) and \(v=\sinh^{-1}(2x)\), then compute the derivatives \(u'=2\) and \(v'\) to find the derivative for the first part of the function. Clear and accurate execution of the product rule is crucial for obtaining the correct derivative, especially in complex functions.

Chain Rule for Derivatives

The chain rule is another indispensable tool in calculus that helps compute the derivative of a composed function. When a function comprises other functions that depend on each other, the chain rule allows us to peel off these layers, step by step, to find the overall derivative.

In mathematical terms, if a function can be written as \(f(g(x))\), then its derivative is \(f'(g(x))\cdot g'(x)\). This rule is particularly useful when differentiating the second part of our exercise's function: \( -\sqrt{1+4x^2} \), which involves a square root, an operation that can be viewed as an outer function \(f(u) = \sqrt{u}\) where \(u=1+4x^2\).

The application of the chain rule leads us to the derivative of this part as \(-\frac{1}{2}(1+4x^2)^{-\frac{1}{2}} \cdot 8x\), where \(8x\) is the derivative of the inner function \(1+4x^2\). It's essential to recognize when and how to apply the chain rule correctly to break down composite functions.

Inverse Hyperbolic Sine Function

The inverse hyperbolic sine function, written as \(\sinh^{-1}(x)\) or 'arsinh(x)', is the inverse of the hyperbolic sine function and appears in various mathematical contexts, including integrals and derivative problems.

This inverse function allows us to solve for angles in hyperbolic trigonometry analogously to the way the inverse sine function does in circular trigonometry. The derivative of \(\sinh^{-1}(x)\) is crucial when dealing with problems involving hyperbolic functions and their inverses.

The derivative is \[\frac{d}{dx}\sinh^{-1}(x)=\frac{1}{\sqrt{1+x^2}}\]. In our exercise, this derivative becomes part of the final result when differentiating the function \(y=2x\sinh^{-1}(2x)\), reflecting the connection between hyperbolic functions and their derivatives in calculus.

Implicit Differentiation

Sometimes, functions are written in forms that are not explicitly solved for one variable in terms of another—these are known as implicit functions. Implicit differentiation is a technique that allows us to find the derivative of such functions by taking the derivative of both sides with respect to the independent variable.

In essence, we treat y as a function of x (even when not isolated) and differentiate both sides accordingly, often applying the chain rule along the way. This technique is particularly useful when direct differentiation isn't possible or would be overly complex.

While the given exercise does not directly involve an implicit function, understanding this concept is beneficial for a comprehensive grasp of differentiation techniques. Implicit differentiation broadens our toolkit for handling a variety of calculus problems beyond the scope of basic derivative rules.