solve for the derivative?

Question

of (ln(t))/t^2?
i did it but dont get the same as the answer in the back of the book.

Answer 1

f(t) = [ln(t)] / [t^2}

To solve this, you have to use the quotient rule. A reminder:
"The derivative of the top times the bottom minus the derivative of the bottom times the top, over the bottom squared."

f'(t) = [ (1/t) t^2 - 2t ln(t) ] / [t^2]^2

Now, simplify as normal.

f'(t) = [ t - 2t ln(t) ] / t^4

Note that we can factor a t out on the top.

f'(t) = t[1 - 2ln(t)] / t^4

Now, we can cancel out the t on the top by reducing the power t^4 by 1.

f'(t) = [1 - 2ln(t)] / (t^3)

Answer 2

(ln t) / t^2

((t^2 * (1/t)) - (ln (t) * 2t)) / (t^4)
(t - 2t ln t) / (t^4)
(t (1 - 2 ln t)) / (t^4)
(1 - 2 ln t) / t^3

This may be written (1 - ln (t^2)) / t^3 or (1 + ln (t^-2)) / t^3 as well.

If you are having problems with finding the derivative of quotients, do you know the mnemonic "lo d hi minus hi d lo over lo lo"? It really helps in remembering how to find those derivatives. . .simply follow this:

d/dx (hi / lo) = ((lo * d hi) - (hi * d lo)) / ho^2

where "d hi" and "d lo" are derivatives of "hi" and "lo" respectively.

Answer 3

derivative of ln(t) is 1/t
derivative of t^2 is 2t

Combine these and use the quotient rule to solve for t

quotient rule = (f/g)'=(f'g-fg')/g^2

Answer 4

Think of it as (ln(t)) X t^-2 and try it again.