# Question #a1919

Jul 5, 2017

$F ' \left(3\right) = 152.$

#### Explanation:

Given that, $F \left(x\right) = 5 {x}^{3} + 4 {x}^{2} - 7 x + 4 ,$ and we have to find $F ' \left(3\right) .$

We know that, $\left({x}^{n}\right) ' = n {x}^{n - 1} \ldots \ldots \ldots \ldots \ldots \ldots . \left(\ast\right) .$

Therefore, we have,

$F ' \left(x\right) = \left\{5 {x}^{3} + 4 {x}^{2} - 7 x + 4\right\} ' ,$

$= \left(5 {x}^{3}\right) ' + \left(4 {x}^{2}\right) ' - \left(7 x\right) ' + \left(4\right) ' ,$

$= 5 \left({x}^{3}\right) ' + 4 \left({x}^{2}\right) ' - 7 \left({x}^{1}\right) ' + 0 ,$

$= 5 \left(3 {x}^{3 - 1}\right) + 4 \left(2 {x}^{2 - 1}\right) - 7 \left(1 {x}^{1 - 1}\right) \ldots \ldots \ldots \ldots \left[\because , \left(\ast\right)\right] ,$

$= 5 \left(3 {x}^{2}\right) + 4 \left(2 {x}^{1}\right) - 7 \left({x}^{0}\right) ,$

$\therefore F ' \left(x\right) = 15 {x}^{2} + 8 x - 7.$

$\therefore F ' \left(3\right) = 15 {\left(3\right)}^{2} + 8 \left(3\right) - 7 = 135 + 24 - 7 ,$

$\Rightarrow F ' \left(3\right) = 152.$