How do you calculate the derivative of f(x) = 1/(x+3)?

Question

It has to be differentiated by forming the difference quotient [f(x+h) - f(x)]/h and taking the limit as h tends to 0.

Any help appreciated!

Answer 1

d/dx of 1/(x+3)
=d/dx of (x+3)^-1
=-1(x+3)^-2
=-1/(x+3)^2

Answer 2

f(x) = 1/(x+3)

[f(x+h) - f(x)]/h =
[1/(x+h+3) - 1/(x + 3)]/h =
[(x + 3 - x - h - 3)/(x + h + 3)(x + 3)]/h =
[-h/(x + h + 3)(x + 3)]/h] =
-1/(x + h + 3)(x + 3)
f'(x) = lim(-1/(x + h + 3)(x + 3)) as h → 0 =
f'(x) = -1/(x + 3)(x + 3) =
f'(x) = -1/(x + 3)^2

Answer 3

f1(x) = limit h to 0 [f(x+h)-f(x)]/h
= limit h to 0 [ 1/(x+h+3)-1/(x+3)]/h
=limit h to 0 [ {(x-3)-(x+h+3)}/(x+h+3)(x+h)]/h
=limit h to o [ -h/{(x+3)^2+4h}{h} ((expanding the denominator and refactorize it).

= limit h to 0 [ -1/(x+3)^2 +4h]
= -1/(x+3)^2

Answer 4

quotient rule (lo d hi - hi d lo)/ lolo
((x+3) * 0 - (1)(1))/ (x+3)^2
1/ (x+3)^2 or 1/x^2+6x+9
i just took a calc test with this on it tuesday, have fun with related rates they suck.

Answer 5

1(X+3)'-(x+3)1 / (x+3)^2
= 1-(x+3) / (X+3)^2

or limit as h go to 0, {[1/((x+h) + 3)] - [1/(x+3)]} / h

Answer 6

d/dx of 1/(x+3)
=d/dx of (x+3)^-1
=(-1)*(x+3)^-2
=(-1)/(x+3)^2