How to find the slope for an equation of the third degree? I need your help.
My question is, how do I find the slope for an equation of the third degree?
For example, if we want to find the slope for an equation of the second degree, we find its first derivative like this:
$$f(x)= x^2 -1 \Rightarrow 
f'(x)= 2x \Rightarrow
f'(1)=2$$
Slope is equal to $2$ in our example. However, I do not know how to find the slope for an equation of the third degree. I need your help!
Clear example with images would be appreciated.
Thanks everybody.
 A: The derivative gives you the slope of a tangent at a given point. So if you want to find the slope of, say $f(x)=x^3-12x-5$ at $x=3$, you would differentiate it once to get $f'(x)=3x^2-12$ and substitute in $x=3$. So the slope of the equation at $x=3$ would be $15$.
It doesn't matter what type of function you have, the first derivative will always give you the slope of the function.
A: $6$ times spelling "slop." $5$ times spelling "fined." I didn't even know browsers without autocorrect/spell-check existed.
To answer your question, you need to learn the general rule to take the derivative of a function. The rule you're looking for is the power rule, which states that, if $\varphi$ is some constant and $u$ is some variable, then:
$$\frac{d}{dx}u^\varphi=\varphi u^{\varphi -1}\frac{du}{dx}$$
In this case, since $u$ is $x$, this can be changed to be:
$$\frac{d}{dx}x^\varphi=\varphi x^{\varphi -1}$$
Hence, when you have $x^2$, its slope would be $2x^{2-1}=2x$. Therefore, for a third power, $x^3$ for instance would be $3x^{3-1}=3x^2$.
