Calculus and Its Applications, Chapter 1, 1.6, Section 1.6, Problem 26

Differentiate $F(x) = (-3x^2 +4x)( 7 \sqrt{x} + 1)$
By multiplying first before differentiating we get

$\begin{equation} \begin{aligned} F(x) &= \left( -3x^2 + 4x \right) \left( 7x^{\frac{1}{2}} + 1 \right)\\ \\ F(x) &= -21x^{\frac{5}{2}} - 3x^2 + 28x^{\frac{3}{2}} + 4x \end{aligned} \end{equation}$


Thus,

$\begin{equation} \begin{aligned} F'(x) &= \frac{d}{dx} \left( -21x^{\frac{5}{2}} - 3x^2 + 28x^{\frac{3}{2}} + 4x \right)\\ \\ &= -21 \cdot \frac{d}{dx} \left( x^{\frac{5}{2}} \right) - 3 \cdot \frac{d}{dx} (x^2) + 28 \cdot \frac{d}{dx} \left( x^{\frac{3}{2}} \right) + 4 \cdot \frac{d}{dx} (x)\\ \\ &= -21 \cdot \frac{5}{2} \left( x^{\frac{5}{2} - 1} \right) -3 \cdot 2 \left( x^{2 - 1} \right) + 28 \cdot \frac{3}{2} \left( x^{\frac{3}{2} - 1} \right) + 4 \cdot (1)\\ \\ &= \frac{-105}{2} x^{\frac{3}{2}} - 6x + 42x^{\frac{1}{2}} + 4\\ \\ &= \frac{-105}{2} \sqrt{x^3} - 6x + 42 \sqrt{x} + 4 \end{aligned} \end{equation}$