Question

If\(f\left( x \right) = \ln \left( {x + \ln x} \right)\),find \({\bf{f'}}\left( {\bf{1}} \right)\).

Short answer

The required answer is \(f'\left( 1 \right) = 2\).

Step-by-step solution

Write the formula of the derivatives of logarithmic functions, the chain rule, and the power rule

Derivative of logarithm function: \(\frac{d}{{dx}}\left( {\ln x} \right) = \frac{1}{x}\),

The Chain Rule:If the function \(g\) is a differentiable function at \(x\) and \(f\) is a differentiable function at \(g\left( x \right)\) the composite function\(F = f \circ g\) defined by \(F\left( x \right) = f\left( {g\left( x \right)} \right)\) is a differentiable function at \(x\) and \(F'\) is given by the product shown below, \(F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\).

Domain: The domain is the set of all the values for which the function is well defined.

Find the differentiation of the function

Consider the function\(f\left( x \right) = \ln \left( {x + \ln x} \right)\). Differentiate the function w.r.t \(x\) by using the derivatives of logarithmic functions and the chain rule.

\(\begin{aligned}{c}\frac{d}{{dx}}\left( {f\left( x \right)} \right)&= \frac{d}{{dx}}\left( {\ln \left( {x + \ln x} \right)} \right)\\&= \frac{1}{{\left( {x + \ln x} \right)}}\frac{d}{{dx}}\left( {x + \ln x} \right)\\&= \frac{1}{{\left( {x + \ln x} \right)}}\left( {1 + \frac{1}{x}} \right)\end{aligned}\)

Find \({\bf{f'}}\left( {\bf{1}} \right)\)

Put \(x = 1\) into \(f\left( x \right) = \frac{1}{{\left( {x + \ln x} \right)}}\left( {1 + \frac{1}{x}} \right)\)

\(\begin{aligned}{c}f'\left( 1 \right)&= \frac{1}{{\left( {1 + \ln 1} \right)}}\left( {1 + \frac{1}{1}} \right)\\&= \frac{1}{{1 + 0}}\left( {1 + 1} \right)\\&= 2\end{aligned}\)

Thus, \(f'\left( 1 \right) = 2\).