Question

state the definition of the derivative and use only the limit definition to evaluate f'(x) for f(x)=3/2x^2+7x-1

Answer 1

The derivative for function f(x) at a specific point x, is the slope of the tangent to f(x) at point x.
To get an estimate for the slope of a line tangent to f(x) near a specific value of x , we look at "rise"/"run" near that x .
"run" means the change in x, and "run" means the change in f(x).
The smaller we make the "run" (as close to zero as possible),
the more accurate our answer will be.

\[\frac{Rise}{Run}=\lim_{a \rightarrow 0} \frac{f(x+a)-f(x)}{(x+a)-x}=\lim_{a \rightarrow 0} \frac{f(x+a)-f(x)}{a}\]

In our case:
\[\lim_{a \rightarrow 0}\frac{(\frac{3}{2}(x+a)^2+7(x+a)-1)-(\frac{3}{2}x^2+7x-1)}{a}\]
\[=\lim_{a \rightarrow 0}\frac{(\frac{3}{2}(x^2+2ax+a^2)+7x+7a-1)-(\frac{3}{2}x^2+7x-1)}{a}\]
\[=\lim_{a \rightarrow 0} \frac{\frac{3}{2}x^2+3ax+\frac{3}{2}a^2+7x+7a-1-\frac{3}{2}x^2-7x+1}{a}\]
\[=\lim_{a \rightarrow 0}\frac{3ax+\frac{3}{2}a^2+7a}{a}=\lim_{a \rightarrow 0} \frac{a(3x+\frac{3}{2}a^2+7)}{a}\]
\[=\lim_{a \rightarrow 0} (3x+7+\frac{3}{2}a)\]
\[=(3x+7)+\lim_{a \rightarrow 0}\frac{3}{2}a\]\[=(3x+7)\]

so, for
\[f(x) = \frac{3}{2}x^2+7x-1\]\[f'(x)=3x+7\]

Answer 2

Small correction to the above:
"run" means the change in x, and "rise" means the change in f(x).