Question

find the derivative of f(x)=3x^2-5x using f'(x)= lim y->x(f(y)-f(x))/y-x

Answer 1

where are you stuck??

Answer 2

Im not sure how to plug it in. I should put it in for x would I? But what do I do with Y?

Answer 3

\[
\lim_{y\to x}\frac{f(y)-f(x)}{y-x} = \lim_{y\to x}\frac{3y^2-5y-(3x^2-5x)}{y-x}
\]

Answer 4

Can you do it from there?

Answer 5

Not really. Do i need to take out an x and y?

Answer 6

You need to simplify a bit.

Answer 7

\[
\lim_{y\to x}\frac{3y^2-5y-(3x^2-5x)}{y-x}
=
\lim_{y\to x}\frac{(3y^2-3x^2)-(5y-5x)}{y-x}
\]

Answer 8

Getting it yet?

Answer 9

couldn't i do it like this \[y(3y-5)-x(3x-5)/ y-x\]

Answer 10

\[
\lim_{y\to x}\frac{(3y^2-3x^2)-(5y-5x)}{y-x}
=
\lim_{y\to x}\frac{3(y^2-x^2)-5(y-x)}{y-x}
\]

Answer 11

\[

\lim_{y\to x}\frac{3(y^2-x^2)-5(y-x)}{y-x}
=

\lim_{y\to x}\frac{3(y-x)(y+x)-5(y-x)}{y-x}
\]

Answer 12

\[

\lim_{y\to x}3(y+x)-5
\]

Answer 13

@ns585 Now do you get it? I had to do difference of squares.

Answer 14

yes i see what you did but don't I have to do something else also or is that the answer?

Answer 15

First of all you need to apply the Sum and Difference rule. that is\[\frac{ d }{ dx } (f(x)-g(x))= \frac{ d}{ dx } (f(x))- \frac{ d }{ dx } (g(x))\]
In your case is\[f(x)=3x^2-5x\]
then we just follow the rule\[f(x)= \frac{ d }{ dx }(3x^2)-\frac{ d }{ dx}(5x)\]
use the power rule for 3x^2 that should be\[n x^n-1\]
at the end it would be\[f(x)=6x-5\]

Answer 16

sorry my bad the power rule is\[n x^{n-1}\]

Answer 17

\[

\lim_{y\to x}3(y+x)-5=3(x+x)-5=6x-5
\]