Asked by Darly

Sorry! I don't get this one either..

limit of (f(x+Δx)-f(x)) / Δ x with Δ x -> 0
f(x)= 1/(x+3)

My work so far before I got stuck...
1/((x+Δx)+3)
= 1/(x+Δx+3) - 1/x+3
= 1(x+3)/(x+Δx+3)(x+3) - 1(x+Δx+3)/(x+3)(x+Δx+3)
=Δx / (x+Δx+3)(x+3)

I need to put Δx/(x+Δx+3)(x+3) divided by Δx, but I don't know how to solve that..

Answered by Damon
oh, much more fun !
Answered by Damon
ALMOST !!! sign error

- dx /[(x+dx+3)(x+3) ]

divide by dx

-1/[(x+dx+3)(x+3) ]

now let dx -->0

-1/(x+3)^2

which any calculus text will tell you is the right answer

d/dx (1/u) = [bottom*d top - top d bottom]/bottom squared
Answered by Darly
OH opps! Thank you!!
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