Struggling in following this problem and it's solution.
Problem:
$f(x, 2) = 8x^2$
Solution:
$\frac{d}{dx}f(x, 2)=\frac{d}{dx}(8x^2)=16x$
$x=3$
$16(3)=48$
I got to the part where it's $8x^2$ - but then assumed the solution would be $8 * 3^2 = 72$
Guess what I am asking is why was the exponent moved down and multiplied instead of being used as an actual exponent?
$f(x,2)=8x^2$ is the function that you want to take the derivative.
We compute the derivative and found
$$\frac{d}{dx} f(x,2) = \frac{d}{dx}(8x^2) = 8(2)x^{2-1}=16x$$
Since you want to compute the derivative, we evaluate the derivative at $x=3$.
We have for a monomial, $x^n$,
$$\frac{d}{dx}x^n = nx^{n-1}$$
If you are interested in the value of a function, $f$, we compute $f(x)$. If we are interested in the derivative, we compute $f'(x) = \frac{d}{dx}f(x)$.
