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Chapter 5: Problem 57

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

Short Answer

Expert verified
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 by step solution

01

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.
02

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 \(\frac{1}{1+x^2}\).
03

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, \((\frac{x}{1+x^{2}})' = \frac{1*(1+x^2) - x*2x}{(1+x^2)^2} = \frac{1}{(1+x^2)^2}\).
04

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, \(y' = \frac{1}{1+x^2} + \frac{1}{(1+x^2)^2}\)

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