Question

What are the rules for removing H in terms of derivatives exactly?

Answer 1

Say I have this equation, \[y = -\frac{ 2 }{ x }\]
why does it equal\[\frac{ 2 }{ x^{2} }\]
instead of "Undefined," when \[\frac{ 2h }{ x(x-h)h }\]

Answer 2

\[\lim_{h \rightarrow 0}\]

Answer 3

@Babynini

Answer 4

Simplify until you reach this point:
\[\frac{ -2x+2x+2h }{ x^2+xh }*\frac{ 1 }{ h }\]
[since it was all over h I just rewrote it so that we wouldn't have such a complex fraction. make sense?]

Answer 5

\[\frac{ 2h }{ x^2+xh^2 }\]

Answer 6

I lied.

Answer 7

I GOT IT.

Answer 8

Alright, watch, dearie.

Answer 9

\[\lim_{h \rightarrow 0} \frac{ -\frac{ 2 }{ x-h } -(- \frac{ 2 }{ x })}{ h }\]

Then you have, \[\frac{ -2x + 2x + 2(h) }{ \frac{ x(x-h) }{ h } }\]

Afterwards, you get \[\frac{ 2(h) }{ x^2-xh(h) }\]

Cancel now, and you end up with \[\frac{ 2 }{ x^2 - xh }\]

as h approaches 0, you end up with \[\frac{ 2 }{ x^2 }\]

Answer 10

Oh you evil genius.

Answer 11

Learned from the best. :P

Answer 12

:')