Question

Find f '(a).
f(t) = 6t + 38/t + 7
Please show step by step..thanks

Answer 1

Okay :) Well I would first rewrite this into a form that seems familiar and easy to work with. Rewrite this as f(t)=6t+38(t^(-1))+7.
From here, apply the power rule to find the derivative.
f'(t)=6+38(-1)(t^(-2)).
f'(t)=6-38/t^2.
From here you can just plug in your a value for t. Rendering: f'(a)=6-38/a^2.

Two points of note if you are confused. The constant(7) goes to zero when you differentiate it. The second is that the power rule says: \[d/dx(x^n)=(n)x^{n-1}\]

Answer 2

(6t)'=6
(38/t)'=(38t^{-1})=-1*38t^{-1-1}=-38t^{-2}=-38/t^2
(7)'=0

Answer 3

f'(t)=6-38/t^2
f'(a)=6-36/a^2