Find f '(t) using the definition of derivative. f(t) = (1-3t)/(2+t)

The definition of derivative of a function  is
  (the limit of the quotient of the difference between f(x + h) and f(x) and h, as h approaches zero.)
To apply this definition to the given function  , fist find f(t + h) by plugging in t+h instead of every t:

Then, find the difference between f(t + h) and f(t):

Simplify the numerator by using distributive property:

Dividing this by h results in

Finally, consider the limit when h approaches zero. Then, the expression above approaches the value of the derivative of f(t):

We can use the quotient rule to solve this derivative.
f(t) = (1-3t)/(2+t)


where g(x) = 1-3t and h(x) = 2+t
f'(t) = [-3(2+t) - 1(1-3t)]/[(2+t)^2]
=-7 /(2+t)^2

The definition of derivative of a function f(x) is
f'(x) = lim_(h ->0) (f(x+h) - f(x))/h  (the limit of the quotient of the difference between f(x + h) and f(x) and h, as h approaches zero.)
To apply this definition to the given function f(t) = (1-3t)/(2+t) , fist find f(t + h) by plugging in t+h instead of every t:
f(t + h) = (1 - 3(t + h))/(2 + (t + h)) = (1 - 3t-3h)/(2 + t + h)
Then, find the difference between f(t + h) and f(t):
f(t+h) - f(t) = (1 - 3t-3h)/(2 + t + h)-(1-3t)/(2+t) = ((1-3t-3h)(2+t) - (1-3t)(2+t+h))/((2+t+h)(2+t))
Simplify the numerator by using distributive property:
f(t+h)-f(t) = ((1-3t)(2+t)-3h(2+t) - (1-3t)(2+t)-h(1-3t))/((2+h+t)(2+t)) =(-6h-3ht-h+3ht)/((2+h+t)(t+h)) = (-7h)/((2+h+t)(t+h))
Dividing this by h results in
(f(t + h)-f(t))/h=(-7)/((2+t+h)(2+t))
Finally, consider the limit when h approaches zero. Then, the expression above approaches the value of the derivative of f(t):
f'(t) = lim_(h->0) (f(t+h)-f(t))/h = -7/(2+t)^2
This result can be confirmed by taking the derivative of f(t) using the quotient rule and the chain rule:
f'(t) = ((1-3t)'(2 + t) - (2+t)'(1-3t))/(2+t)^2 = (-3(2+t) - 1(1-3t))/(2+t)^2 = -7/(2+t)^2