# How do I solve this derivative at one point?

## THANK YOU

Feb 17, 2017

$F ' \left(x\right) = 3$

#### Explanation:

to find the derivative of F(x) you use the Power Rule which is $\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1}$ and the Constant Rule which is $\frac{d}{\mathrm{dx}} \left(c\right) = 0$

$\frac{d}{\mathrm{dx}}$ just means to take the derivative in terms of x, $\frac{d}{\mathrm{dx}} = F ' = F ' \left(x\right)$

Power Rule: $\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1}$
Constant Rule: $\frac{d}{\mathrm{dx}} \left(c\right) = 0$
(there are more derivative rules)

$F \left(x\right) = 3 x - 5$
$\frac{d}{\mathrm{dx}} F \left(x\right) = \frac{d}{\mathrm{dx}} \left(3 x - 5\right)$
$F ' \left(x\right) = 3 {x}^{1 - 1} + 0$
*(5 becomes 0 because of the Constant Rule and 3x becomes 3 because of the Power Rule)

$F ' \left(x\right) = 3$
now we are supposed to plug in $x = - \frac{3}{4}$ into the derivative to get a y-value... but you can't because it's just a constant...

(Note: $\frac{d}{\mathrm{dx}}$ "means taking the derivative with respect to x" which means I want the derivative to contain x-variables. You might see $\frac{\mathrm{dx}}{\mathrm{dy}} \mathmr{and} \frac{d}{\mathrm{dy}}$ which means to "take the derivative with respect to y", taking the derivative so that there are only y-variables...

For example, there might be an equation that looks like this:
$V = 2 x + {x}^{2}$ and they ask you to find $\frac{d}{\mathrm{dx}}$ so you just find the derivative and since all the variables are x you are good. Now, if they want you to find $\frac{d}{\mathrm{dy}}$ then they want to find the derivative and they want the all the x-variables to be y-variables. You won't have to worry about that later when you learn the Chain Rule.)