# How do you find the slope of the graph f(x)=(2x+1)^2 at (0,1)?

Using the chain rule of derivatives, and then plugging in a x value of 0 for the $f ' \left(x\right)$ function, you get a slope.

#### Explanation:

Consider
$F \left(x\right) = f p r i m e \left(g \left(x\right)\right) \cdot g p r i m e \left(x\right)$ This what we define as the chain rule, "first the outside derivative then the inside derivative"
If $f \left(x\right) = {\left(2 x + 1\right)}^{2}$
and $f ' \left(x\right) = 8 x + 4$
and when $f ' \left(0\right) = 8 \left(0\right) + 4 = 4$
Then the slope equals 4.

Proof Thing:
Derivatives are slopes of functions.
We know this by the formal definition of $f ' \left(x\right) = \frac{f \left(x + h\right) - f \left(x\right)}{h}$
and $m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$
They are similar because that is how the definition came by substituting them as $\frac{\Delta y}{\Delta x}$; where $h$ represents the change the function.
So, derivatives find a slope of a function.
We simplify the process by using various derivative rules