How do you use the Product Rule to find the derivative of g(x) = (2x^2 + 4x - 3) ( 5x^3 + 2x + 2)?

Aug 16, 2015

${g}^{'} = 50 {x}^{4} + 80 {x}^{3} - 33 {x}^{2} + 24 x + 2$

Explanation:

Notice that your function can be written as the product of two other functions

$g \left(x\right) = {\underbrace{\left(2 {x}^{2} + 4 x - 3\right)}}_{\textcolor{\mathmr{and} a n \ge}{f \left(x\right)}} \cdot {\underbrace{\left(5 {x}^{3} + 2 x + 2\right)}}_{\textcolor{p u r p \le}{h \left(x\right)}}$

This means that you can rule the product rule to get

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(g \left(x\right)\right) = \left[\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right)\right] \cdot h \left(x\right) + f \left(x\right) \cdot \frac{d}{\mathrm{dx}} \left(h \left(x\right)\right)}$

Using this formula will get you

$\frac{d}{\mathrm{dx}} \left(g \left(x\right)\right) = \left[\frac{d}{\mathrm{dx}} \left(2 {x}^{2} + 4 x - 3\right)\right] \cdot \left(5 {x}^{3} + 2 x + 2\right) + \left(2 {x}^{2} + 4 x - 3\right) \cdot \frac{d}{\mathrm{dx}} \left(5 {x}^{3} + 2 x + 2\right)$

${g}^{'} = \left(4 x + 4\right) \cdot \left(5 {x}^{3} + 2 x + 2\right) + \left(2 {x}^{2} + 4 x - 3\right) \cdot \left(15 {x}^{2} + 2\right)$

${g}^{'} = 20 {x}^{4} + 8 {x}^{2} + 8 x + 20 {x}^{3} + 8 x + 8 + 30 {x}^{4} + 4 {x}^{2} + 60 {x}^{3} + 8 x - 45 {x}^{2} - 6$

${g}^{'} = \textcolor{g r e e n}{50 {x}^{4} + 80 {x}^{3} - 33 {x}^{2} + 24 x + 2}$