Question

[3(-3+h)^2 + 2(-3+h)-21]/h

Answer 1

derivative :?

Answer 2

yeah the whole problem is a derivative, but i can't multiply this part, keep getting the wrong answer.

Answer 3

ok

i may can help, is that the original equation problem :?

Answer 4

the original is f'(-3)= Lim h-->0 [f(-3+h)-f(-3)]/h and f(x)= 3x^2 +2x

Answer 5

ok, lemme give you the equation as i worked it and see if it will help,

the derivative formula i used was (f(x+h)-f(x))/h

Answer 6

so you start with your given F(x)
3X^2+2x

now for the derivative formula you plug (X+h) into the X position

giving you 3(x+h)^2 + 2(x+h)

then add the - F(x) for a long problem of
3(x+h)^2 + 2(x+h) - (3x^2+2x)

you distribute and get

3(x^2 +2xh+ h^2) + 2x + 2h - (3x^2+2x) **as a side note if your writing this every step needs to be over H

then finish the distribution for (3x^2 + 6xh + 3h^2 +2x+2h -3x^2-2x) all over H

then you cancel out the 3x^2 with - 3x^2 the +2x with - 2x and

\[(6xh+3h^2+2h)\h\]

Answer 7

its a bit long so from the (6xh+3h^2+2h)\h you use division rules to cancel one h from the denominator and numberator for 6x + 3h +2

h=0

so your derivative should be 6x+2

Answer 8

Aha I see now, it's much clearer now how you plug in the f(x) into the derivative formula

Answer 9

thank you very much1

Answer 10

yes, i just conquered this myself a few days ago, and as always (once you learn it) use the power rule to check your answer

Answer 11

power rule to check, i'll remember that. thx