Question

find derivatives using the limit definition ?i am stuck in the final steps?

Answer 1

2x^2-16x+35
useing this
\lim_{h \rightarrow 0} \frac{ f(x+h)-f(x) }{ h }

Answer 2

\[ \lim_{h \rightarrow 0} \frac{ f(x+h)-f(x) }{ h }\]

Answer 3

What's the hard part?

Answer 4

so i did it like this \[\lim_{h \rightarrow 0} \frac{2(x+h)^2-16(x+h)+35 }{ h }\] and then \[\lim_{h \rightarrow 0} \frac{2(x^2+2xh+h^2)-16x-16h+35 }{ h }\] and lastly \[\lim_{h \rightarrow 0} \frac{2x^2+4xh+4h^2-16x-16h+35 }{ h }\]

Answer 5

where is f(x) ?

Answer 6

\[f(x)=2x^2-16x+35\]

Answer 7

you forgot \(\dfrac{f(x+h)\color{red}{\LARGE{-f(x)}}}h\)

Answer 8

\(\large \lim_{h \rightarrow 0} \frac{2(x+h)^2-16(x+h)+35 - \color{red}{f(x)} }{ h }\)

Answer 9

\[\lim_{h \rightarrow 0}\frac{ [2(x+h)^2-16(x+h)+35]-[2x^2-16x+35] }{ h }\]

Answer 10

ohhh nooo i forgot to do it, thanks @kc_kennylau

Answer 11

no problem :)

Answer 12

\[\lim_{h \rightarrow 0}\frac{ [2(x^2+2xh+h^2)-16x-16h+35]-[2x^2-16x+35] }{ h }\]

Answer 13

\[\lim_{h \rightarrow 0}\frac{ 2x^2+4xh+2h^2-16x-16h+35-2x^2+16x-35 }{ h }\]

Answer 14

\[\lim_{h \rightarrow 0}\frac{ 4xh+2h^2-16h }{ h}\]

Answer 15

Is it so freaking difficult to learn to type \(LaTeX\)? :o

Answer 16

\[\lim_{h \rightarrow 0}\frac{ 4x+2h-16 }{ }\]

Answer 17

I meant \(\LaTeX\)*

Answer 18

\[\lim_{h \rightarrow 0}4x+2(0)-16\]

Answer 19

\[=4x-16\]

Answer 20

And yeah, It is fluttering difficult to type it, you got a problem with that?

Answer 21

feather.

Answer 22

xDDD no offffffffense

Answer 23

13itch

Answer 24

xD

Answer 25

calm down bro, i was just kidding :P

Answer 26

Thank you very much @FutureMathProfessor for taking the time to show me the steps .

Answer 27

you can't be a future math professor if you don't calm down :)

Answer 28

anytime @boyb39 <3