Find the instantaneous velocity

Function: f(x) = 1 / (x + 1)
Interval: [0,3]

The answer in the book is f ' (0) = -1 and f ' (3) = - 1/16. I couldn't get the answer.

f(x) = 1 / (x + 1) = (x + 1)^-1

f ' (x) = -1(x +1)^-2 = -1 / (x + 1)^2

f '(0) = -1 / (0 + 1)^2 = -1/1 = -1

f '(3) = -1 / (3 + 1)^2 = -1/(4)^2 = -1/16

you have to find the derivative of the function
remember you can use either the product rule or the quotient rule

i used the quotient rule
easy way to remember
LOW D HIGH - HIGH D LOW... DRAW A LINE AND SQUARE BELOW!

doing this you get:
f ' (x) = (x+1)(0) - 1(1) / (x+1)²
simplify
f ' (x) = -1 / (x+1)²
now you have [0,3] for your interval.. just plug each into the derivative you found
for 0:
f ' (0) = -1 / (0+1)²
f ' (0) = -1/1
f ' (0) = -1
for 3:
f ' (3) = -1 / (3+1)²
f ' (3) = -1 / 4²
f ' (3) = -1/16

The derivative of f(x) is -1/(x+1)^2. Plug in x = 0 and x = 3 and you'll get those answers.

How do you get that derivative? f(x) = (x + 1)^(-1) so by the power rule and the chain rule, f'(x) = -1(x+1)^(-2) * d/dx(x + 1)