To differentiate the given expression, \(\frac{5x^3 + x^2}{x}\), we can apply the quotient rule.

The quotient rule states that for a function of the form \(\frac{u(x)}{v(x)}\), the derivative is calculated as:

\(\left(\frac{u(x)}{v(x)}\right)' = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\)

Let's differentiate the given expression step by step:

Given function: \(\frac{5x^3 + x^2}{x}\)

Applying the quotient rule:\(\left(\frac{5x^3 + x^2}{x}\right)' = \frac{(3(5x^2) + 2x)(x) - (5x^3 + x^2)(1)}{(x)^2}\)

Simplifying the expression:\(\left(\frac{5x^3 + x^2}{x}\right)' = \frac{(15x^2 + 2x)(x) - (5x^3 + x^2)}{x^2}\)

Expanding and combining like terms:\(\left(\frac{5x^3 + x^2}{x}\right)' = \frac{15x^3 + 2x^2 - 5x^3 - x^2}{x^2}\)

Simplifying further:\(\left(\frac{5x^3 + x^2}{x}\right)' = \frac{10x^3 + x^2}{x^2}\)

Finally, we can simplify the expression by factoring out \(x\) from the numerator:\(\left(\frac{5x^3 + x^2}{x}\right)' = \frac{x(10x^2 + x)}{x^2}\)

Canceling out one \(x\) from the numerator and denominator, we get:\(\left(\frac{5x^3 + x^2}{x}\right)' = \frac{10x^2 + x}{x}\)

Simplifying further:\(\left(\frac{5x^3 + x^2}{x}\right)' = 10x + 1\)

Therefore, the derivative of \(\frac{5x^3 + x^2}{x}\), where \(x \neq 0\), with respect to x is 10x + 1.