Find the derivative of x³ + 2x² − 5x.

Step-by-step solution

1. 1

   Differentiate term by term. Differentiation is linear, so a sum can be handled one term at a time and any constant multiple simply rides along:

   $\frac{d}{dx}\bigl(u + v\bigr) = \frac{du}{dx} + \frac{dv}{dx}, \qquad \frac{d}{dx}\bigl(cu\bigr) = c\,\frac{du}{dx}$

2. 2

   Apply the power rule to the cubic term. The power rule is

   $\frac{d}{dx}x^n = n x^{n-1}$

   so

   $\frac{d}{dx}x^3 = 3x^2$

   The exponent moves out front and drops by one — it does not stay at $3$.

3. 3

   Handle the quadratic term with its coefficient.

   $\frac{d}{dx}\left(2x^2\right) = 2 \cdot 2x = 4x$

   The $2$ out front is untouched by the power rule; only the $x^2$ is differentiated, giving $2x$, and the coefficients multiply to $4$.

4. 4

   Handle the linear term. With $n = 1$ the power rule gives $\frac{d}{dx}x = 1$, so

   $\frac{d}{dx}(-5x) = -5$

   A linear term always differentiates to its own coefficient — and if there had been a constant term, it would have differentiated to $0$.

5. 5

   Assemble and check. Adding the three pieces:

   $f'(x) = 3x^2 + 4x - 5$

   Numeric check with a central difference at $x = 2.5$: $\frac{f(2.5 + h) - f(2.5 - h)}{2h} \approx 23.75$ for small $h$, and $3(2.5)^2 + 4(2.5) - 5 = 18.75 + 10 - 5 = 23.75$. As a bonus, $f'(x) = 0$ factors as $(3x - 5)(x + 1) = 0$, so the cubic has turning points at $x = 5/3$ and $x = -1$.