Question #1092f

Feb 23, 2015

The answer is: $h ' \left(x\right) = {\left(2 x + 7\right)}^{3} \left(x - 1\right) \left(14 {x}^{2} + 11 x - 7\right)$.

$h ' \left(- 3\right) = 344$

Rememering that:

$y = f \left(x\right) \cdot g \left(x\right) \Rightarrow y ' = f ' \left(x\right) \cdot g \left(x\right) + f \left(x\right) g ' \left(x\right)$

Since:

$h \left(x\right) = x {\left(2 x + 7\right)}^{4} {\left(x - 1\right)}^{2}$

than:

$h ' \left(x\right) = 1 \cdot {\left(2 x + 7\right)}^{4} \cdot {\left(x - 1\right)}^{2} + x \cdot 4 {\left(2 x + 7\right)}^{3} \cdot 2 \cdot {\left(x - 1\right)}^{2} + x {\left(2 x + 7\right)}^{4} \cdot 2 {\left(x - 1\right)}^{1} \cdot 1 =$

$= {\left(2 x + 7\right)}^{3} \left(x - 1\right) \cdot \left[\left(2 x + 7\right) \left(x - 1\right) + 8 x \left(x - 1\right) + 2 x \left(2 x + 7\right)\right] =$

$= {\left(2 x + 7\right)}^{3} \left(x - 1\right) \cdot \left(2 {x}^{2} - 2 x + 7 x - 7 + 8 {x}^{2} - 8 x + 4 {x}^{2} + 14 x\right) =$

$= {\left(2 x + 7\right)}^{3} \left(x - 1\right) \left(14 {x}^{2} + 11 x - 7\right)$.

$h ' \left(- 3\right) = {\left(2 \cdot \left(- 3\right) + 7\right)}^{3} \left(- 3 - 1\right) \left(14 \cdot {\left(- 3\right)}^{2} + 11 \cdot \left(- 3\right) - 7\right) =$

$= {\left(1\right)}^{3} \left(- 4\right) \left(86\right) = 344$