Find the derivative of $$f(x)=\dfrac{(x-1)^2}{x}$$.
Development:
The function is a quotient between two functions. Nevertheless, we cannot derive the dividend (it is not one of the elementary functions we know how to deal with). What can be seen, nevertheless, is that we can write it as a product. In general, the function $$f(x)=\dfrac{g(x)}{h(x)}$$, identifying terms,
$$g(x)=(x-1)^2 \, \ \ h(x)=x$$
Before applying the rule of the quotient, I need to know how much will be $$g'(x)$$. Since we have that $$g(x)=(x-1)\cdot(x-1)$$ and we can use the rule of the product to find its derivative. $$$g'(x)=1(x-1)+(x-1)1=2(x-1)$$$ By substituting in the main function: $$$f'(x)=\dfrac{2(x-1)x-(x-1)1}{x^2}=\dfrac{(x-1)(2x-1)}{x^2}$$$
Solution:
$$f'(x)=\dfrac{2(x-1)x-(x-1)1}{x^2}=\dfrac{(x-1)(2x-1)}{x^2}$$