Question

1/h(1/2=h - 1/2)
How do I solve this? The answer is -1/4

Answer 1

what's the whole problem?

Answer 2

Find the derivative of f(x) = 1/x at the point x=2

Answer 3

\[\huge \lim_{h \rightarrow 0} \frac{ \frac{ 1 }{ x+h } -\frac{ 1 }{x } }{ h }\]

Answer 4

you started there...

Answer 5

found a CD in the numerator... ( x(x+h) )

Answer 6

ok this algebra is what I am uncertian of

Answer 7

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

Answer 8

ok that part right there very confusing I have to try to digest it

Answer 9

=-h/(xh(x+h))

Answer 10

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

Answer 11

where did x-(x+h) come from?

Answer 12

1/(x+h)−1/x

what's the CD?

Answer 13

common denominator

Answer 14

x(x+h)

Answer 15

(x / x) *(1 / (x+h)) = x / (x(x+h))

((x+h) / (x+h)) * (-1 / x) = -(x+h) / (x(x+h))

Answer 16

I dont see how to go from x/ (x(x+h) to
((x+h)/(x+h))*(-1/x)

Answer 17

\[\frac{ 3 }{4 } + \frac{ 5 }{3 } = \frac{ 3*3 +4*5 }{ 4*3 }\]

Answer 18

we're finding a CD between those two terms in the numerator.
\[\frac{ 1 }{x+h } - \frac{ 1 }{x } = \frac{ x }{x(x+h) } - \frac{ x+h }{(x+h) x}\]