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..
3 answers
oh, much more fun !
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
- 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
OH opps! Thank you!!
1 answer