What is the 1-st and 2-nd derivative?

[cvik]

1/x+3

[Math Tutor]

Here are first derivative formulas.
f(x) = 1/x + 3
f'(x) = (1/x)' + (3)' = (x^-1)' + 0 =

the derivative of every number is 0

1/x = x^-1
here we use the formula:
y = xⁿ => y' = nx^n-1

The first derivative of x^-1 = -1x^-1-1 = -x^-2

so f'(x) = -x^-2

f''(x) = (f'(x))' = 2x^-3

[cvik]

but in f(-2) the result is -1 for the 1-st derivative and
for the 2-nd derivative f(-2) result is 1?

[cvik]

i finished my calculation and i got
-1/(x+3)² for the 1-st derivative which is exactly -1 for x(-2)

[cvik]

i used the formula (C/V)' = CV'/V²
where C=1 V=x+3

[cvik]

got the 2-nd: 2/(x+3)³
with the same formula CV'/V²
anyway thanks teacher

[Math Tutor]

Sorry but
1/x + 3 is different from 1/(x+3)

[Math Tutor]

f(x) = 1/(x+3)
f'(x) = (1'.(x+3) - 1.(x+3)')/(x+3)²


1'.(x+3) = 0

1.(x+3)' = 1

so f'(x) = -1/(x+3)²

f''(x) = -(-(x+3)²)' / (x+3)⁴
=2(x+3)(1+0)/(x+3)⁴ = 2/(x+3)³


Your answers are correct!

[cvik]

yep my mistake 1/(x+3) problem solved
thanks again