f(t) = 6/t - 8/t^2
find the derivative
i got
-6/t^2 + 16t/t^4
not sure if thats right
thanks!
2007-03-11 20:34:33 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
and im wondering
is the deriv. of root2 -> = 0?
2007-03-11 20:40:24 · update #1
f(t) = 6/t - 8/t^2
One thing you may want to do prior to taking the derivative is to move the variables to the top as a negative exponent.
f(t) = 6t^(-1) - 8t^(-2)
The reason for doing this is to use the power rule with ease.
f'(t) = (-6)t^(-2) + 16t^(-3)
Making the powers positive again, we move them to the bottom.
f'(t) = (-6)/t^2 + 16/t^3
2007-03-11 20:42:23 · answer #1 · answered by Puggy 7 · 0⤊ 0⤋
First term correct. Second should be 16/t^3. Derivative of any constant is 0.
2007-03-11 20:43:57 · answer #2 · answered by Anonymous · 0⤊ 0⤋
f(t) = 6.t^(-1) - 8.t^(- 2)
f ` (t) = - 6 t^(- 2) + 16t^(- 3)
f ` (t) = - 6 / t ² + 16 / t ³
2007-03-11 22:30:31 · answer #3 · answered by Como 7 · 0⤊ 0⤋