Question

Finding the derivative of f(x) the long way:
f(x)=1/x^2; a=1
formula: lim f(x+h)-f(x)/h as h approaches 0.

Answer 1

I'm having difficulties solving it the long way. This is the problem

\[f(x)=\frac{ 1 }{ x^2 };a=1\]
Limit derivative Formula:
\[\lim_{h \rightarrow 0}=\frac{ f(x+h)-f(x) }{ h }\]

Answer 2

I know the answer should be -2/x^3 using the shortcut method, but i'm having difficulty getting to that answer. Here's what I have so far:
\[f(x+h)=\frac{ 1 }{ (x+h )^2}\]
\[f(x) = 1/x^2\]
values plugged into limit formula:
\[\lim_{h \rightarrow 0}\frac{ \frac{ 1 }{ (x+h)^2 }-\frac{ 1 }{ x^2 } }{ h }\]

Answer 3

write the top as a single fraction, simply and expand

Answer 4

so what i had earlier becomes
\[=\lim_{h \rightarrow 0}\frac{ \frac{ 1 }{ x^2+2xh+h^2 }- \frac{ 1 }{ x^2 }}{h }\]

Answer 5

don't expand until after getting the numerator all as one fraction

Answer 6

at least i wouldn't recommend it

Answer 7

i guess it doesn't matter...

Answer 8

@TuringTest
So following your example I get this so far:
\[\lim_{h \rightarrow 0}\frac{ \frac{ 1 }{ (x+h)^2 }-\frac{ 1 }{ x^2 } }{h }=\frac{ \frac{ x^2 -(x+h)^2}{ x^2(x+h)^2 } }{ h }\]

Answer 9

yes, now simplify the denominator and expand the numerator

Answer 10

by 'simplify the denominator' i mean get that triple fraction into a normal one by putting that bottom h in the denom of the upper fraction

Answer 11

Don't you mean in the numerator of the upper fraction, aren't I multiplying the entire upper fraction by 1/h?

Answer 12

i think we mean the same thing, yeah

Answer 13

\[\lim_{h \rightarrow 0}\frac{ \frac{ 1 }{ (x+h)^2 }-\frac{ 1 }{ x^2 } }{h }=\lim_{h\to 0}\frac{ \frac{ x^2 -(x+h)^2}{ x^2(x+h)^2 } }{ h }=\lim_{h\to0}\frac{ x^2 -(x+h)^2}{ x^2(x+h)^2 } \cdot\frac1h\]

Answer 14

ok good thats what I thought.

Answer 15

The above becomes:
\[\lim_{h \rightarrow 0}=\frac{ hx^2-h(x+h)^2 }{ x^2(x+h)^2 }\]
right?

Answer 16

erm.. how did h get into the numerator?

Answer 17

\[\frac ab\cdot\frac cd={ac\over bd}\]

Answer 18

oh ok. I see, so:
\[\lim_{h \rightarrow 0}=\frac{ x^2-(x+h)^2 }{ hx^2(x+h)^2 }\]

Answer 19

yes, now expand the numerator, simplify, and take the limit

Answer 20

\[\lim_{h \rightarrow 0}\frac{ x^2-x^2+2xh+h2 }{ hx^2(x+h)^2 }=\frac{ h(2x+h) }{hx^2(x^2+2xh+h^2)} =\frac{ h(2x+h) }{ hx^4+2x^3h^2+h^3x^2 }\]
\[=\frac{ h(2x+h) }{ hx(x^3+ 2x^2h+h^2x)} \]
this is what I have so far

Answer 21

wonderful, now cancel what can cancel

Answer 22

oh you lost your negative sign, btw

Answer 23

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

Answer 24

\[=\frac{ -2xh-h^2 }{ hx^2(x^2+2xh+h^2) }=\frac{ -2 }{ hx^2(x^2) }=\frac{ -2 }{ x^4h }\]

I think I made a mistake somewhere, cause evaluating at h->0 would give me an undefined answer.