Problem 57 - Calculus of a Single Variable

Question

Find the derivative of the function. \(y=\arctan x+\frac{x}{1+x^{2}}\)

Step-by-Step Solution

Answer

The derivative of the function \(y=\arctan x+\frac{x}{1+x^{2}}\) is \(y' = \frac{1}{1+x^2} + \frac{1}{(1+x^2)^2}\)

Step 1: Identify the Components of the Function

The function \(y=\arctan x+\frac{x}{1+x^{2}}\) is composed of two parts, \(u = \arctan x\) and \(v = \frac{x}{1+x^{2}}\). The derivative of y would be obtained by taking the derivative of both parts \(u\) and \(v\) separately.

Step 2: Take the Derivative of the First Component

Using the formula for the derivative of \(\arctan x\), which is \(\frac{1}{1+x^2}\), the derivative of \(u = \arctan x\) is \(\frac{1}{1+x^2}\).

Step 3: Take the Derivative of the Second Component

The second component of the function is \(v = \frac{x}{1+x^{2}}\), which looks like a fraction, let's use the quotient rule to differentiate it. According to the quotient rule, \((\frac{u}{v})' =\frac{u'v - uv'}{v^2}\). Now, let our \(u = x\) and \(v = 1+x^2\), so \(u' = 1\) and \(v' = 2x\). Plug \(u, v, u'\) and \(v'\) into the quotient rule we get, \((\frac{x}{1+x^{2}})' = \frac{1*(1+x^2) - x*2x}{(1+x^2)^2} = \frac{1}{(1+x^2)^2}\).

Step 4: Combine the Derivatives

The derivative of the original function \(y=\arctan x+\frac{x}{1+x^{2}}\) is obtained by adding the derivatives of its parts. So, \(y' = \frac{1}{1+x^2} + \frac{1}{(1+x^2)^2}\)