Step 1:

We notice that the given function is a product of two other functions. To find the derivative of this, we can use the product rule. First, let’s label the two functions as:

[latex]\large{h(x) = e^x-x}[/latex] and [latex]\large{g(x) = \ln(x)+1}[/latex]

To use the product rule, we also need the derivative of these functions. This should be easy for us to do using our knowledge of common derivatives. Thus:

[latex]\large{h'(x) = e^x-1}[/latex] and [latex]\large{g'(x) = \frac{1}{x}}[/latex]

Step 2:

The product rule formula states that:

[latex]\small{\frac{d}{dx}f(x)g(x) = f(x)g'(x) + f'(x)g(x)}[/latex]

Thus, substituting into the product rule formula, we have:

[latex]\large{\frac{d}{dx}((e^x-x)(\ln(x)+1)) =}[/latex]

[latex]\large{ (e^x-x)(\frac{1}{x}) + (e^x-1)(\ln(x)+1)}[/latex]

Step 3:

Simplifying, we have:

[latex]\large{ = \frac{e^x-x}{x} + (e^x-1)(\ln(x)+1)}[/latex]

We will leave the second term as is, since multiplying it out will make the final solution messier.

Thus, this is the final answer for the derivative. So

[latex]\large{f'(x) = \frac{e^x-x}{x}+(e^x-1)(\ln(x)+1)}[/latex]

Solution Complete!