Step 1:

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

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

To use the quotient rule, we also need the derivative of these functions. We can easily find the derivatives using our knowledge of common derivatives:

[latex]\large{h'(x) = \frac{1}{x}}[/latex] and [latex]\large{g'(x) = -\csc^2(x)}[/latex]

Step 2:

The quotient rule formula states that:

[latex]\large{\frac{d}{dx}\frac{f(x)}{g(x)} = \frac{g(x)f'(x) – f(x)g'(x)}{g(x)^2}}[/latex]

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

[latex]\large{\frac{d}{dx}(\frac{\ln(x)-1}{\cot(x)})}[/latex]

[latex]\small{= \frac{(\cot(x))(\frac{1}{x}) \:- \:(\ln(x)-1)(-\csc^2(x))}{\cot^2(x)}}[/latex]

Step 3:

Simplifying, we have:

[latex]\Large{ = \frac{\frac{\cot(x)}{x}\:+\:+\csc^2(x)\ln(x)\: – \:\csc^2(x)}{\cot^2(x)}}[/latex]

Then factoring out the common factor from the second and third terms, we have:

[latex]\Large{ = \frac{\frac{\cot(x)}{x}\:+\:\csc^2(x)(\ln(x)-1)}{\cot^2(x)}}[/latex]

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

[latex]\Large{f'(x) = \frac{\frac{\cot(x)}{x}\:+\:\csc^2(x)(\ln(x)-1)}{\cot^2(x)}}[/latex]

Solution Complete!