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f(t)=1/t^2-1/t+t

2007-06-02 19:43:16 · 4 answers · asked by Anonymous in Science & Mathematics Mathematics

Show Working Please I am lost?!?

2007-06-02 19:50:08 · update #1

4 answers

f(t) = 1/t^2 - 1/t + t
differentiating both sides wrt t
f'(t) = -2/t^3 + 1/t^2 + 1
= (t^3 + t - 2)/t^3

note:
a) d/dt(1/t^2) = d/dt(t^(-2)) = -2t^(-2-1) = -2t^(-3) = -2/t^3
b) d/dt(1/t) = d/dt(t^(-1)) = -t^(-1-1) = -t^(-2) = -1/t^2
Both according to the formula d/dx(x^n) = nx^(n-1)

2007-06-02 20:00:50 · answer #1 · answered by raja 3 · 0 0

First simplify the fractions using the property (1/x^n = x^-n)

f(t) = t^-2 - t^-1 + t (now find the derivative of each)

f'(t^-2) = -2t^-3
f'(-t^-1) = t^-2
f'(t) = 1

the answer is ...
-2t^-3 + t^-2 +1 or

(in fraction form) -2/t^3 + 1/t^2+1

Both answers are correct.

Also, I answered a previous question of yours very similar to this one, just use some of the properties I explained to you in that question and the ones given to you by your professor and finding the derivative will be much easier for ya! Good luck!

2007-06-03 02:55:35 · answer #2 · answered by surfing86 2 · 0 0

1/t^2 is just like t^-2, so write the equation as:

f(t) = t^-2 - t^-1 + t

Then apply the normal rule of

d(t^a)/dt = at^(a-1)

to get

f'(t) = -2t^-3 + t^-2 + 1

2007-06-03 02:54:04 · answer #3 · answered by Horatio 3 · 0 0

As written,question reads:-
f(t) = t^(-2) - 1.t^(-1) + t
f `(t) = (-2).t^(-3) + 1.t^(-2) + 1
f `(t) = (-2) / t³ + 1/ t² + 1

2007-06-03 13:25:12 · answer #4 · answered by Como 7 · 0 0

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