Can you explain this math problem to me f(t)=3^(2t)/t?

Question

Oops I guess I didnt type that I want you to explain how you would find the derivative of that

Answer 1

To find the derivative, write 3^(2t) as
e^(2t ln3),
then use the product formula
(u'v - uv')/v^2 with

u = e^(2t ln3); v = t

This, of course, gives
((2 ln3)e^(2t ln3)*t - e^(2t ln3))/t^2

= (2t ln3)*3^(2t) - 3^(2t))/t^2, and then we can take out 3^(2t) as a common factor and get
(3^(2t))(2t ln3 - 1)/t^2

It's possible to fiddle around with this a bit more but I can't see it producing any significant simplification. This is it.

Looks like this function has just one stationary point, where
t = 1/(2 ln3), which of course is 1/ln 9.

P.S. Did you understand 3^(2t) = e^(2t ln 3)? It's because e^x and ln x are inverses of each other, so that ln (e^x) and e^(ln x) are each equal to x.
And one of the log properties is that 2t ln3 = ln(3^2t), and so
e^(2t ln3) = 3^(2t).

It's an example of the general formula
a^x = e^(x ln a), from which we find that
the derivative of a^x is (a^x) ln a

Answer 2

It's a function, taking on different values depending on the input. I suppose you mean f(t) = (3^2t)/t, since if you grouped it the other way, 3^(2t/t) it would simplify to 3^2,.which is a constant, 9.
To get an idea as to what this function is doing, you could graph it on a graphics calculator or by hand, or you could plug in numbers for t and see what values it outputs.
First off, you'd notice that t can take on any value, positive or negative, except 0. Substituting some convenient values for t you'd get
(t, f(t)) = (-2, -0.02469...). (-1, -.1111...), (1,9), (2, 40.5), etc
So you see it behaves somewhat like an exponential function only not quite because of the t in the denominator.
There are links online where you can type in a function and see what its graph looks like.
To find the derivative, use the chain rule and the quotient rule. that is, (t(3^2t)' - (3^(2t))(t'))/(t^2)
To take the derivative of (3^2t) make use of the fact that this is equal to e^(2ln3 t) so the derivative would be 2ln3 e^(2ln3 t)
Substitute this back into the quotient rule and you get
(2ln3t(e^2ln3t) - 3^(2t))/t^2 which simplifies to
(2ln3t(3*(2t)) - 3^(2t))/(t^2) =
((3^(2t)(2 ln 3t - 1)) / t^2

Answer 3

f(t) means 'a function of t'
So what you have written means that 'f is a function of t which raises 3 to the power of 2t and then divides this by t'

If, for example, t=1, then f(t) = 3^(2*1)/1 = 3^2 = 9

Ok, to find the derivative, first let y = f(t) so:

y = [3^(2t)]/t
yt= 3^2t

Now differentiate on both sides:

y + ty' = 2[3^(2t)]

So y' = {2[3^(2t)] - y}/t

But y = [3^(2t)]/t so we now have:

y' = {2[3^(2t)] - [3^(2t)]/t }/t which we can simplify to:

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

Hope that is clear enough ;)

Answer 4

Can you explain this math problem to me f(t)=3^(2t)/t

ie f(t) = (3^2)^t / t = 9^t / t

So

t .. | -1.8 | -1.6 | -1.4 | -1.2 |.. -1 ..| -0.1 | -0.01|.. 0 ..|
f(t) |-0.01|-0.02|-0.03|-0.06|-0.11 | -8.0 | -97.8| -∞ |

t .. | 0 | 0.01 | 0.1 | 1/ln9| 1 |. 1.2 | 1.4 | 1.6 | 1.8 |
f(t) |∞|102.2|12.4|. 5.97| 9 | 11.7|15.5|21.0|29.0|

So:
as t → -∞, f(t) →0 (from below)

as t → 0 from below, f(t) → -∞

as t → 0 from above, f(t) → +∞

as t → ∞, f(t) →∞

Now f(t) = 9^t / t =e^(ln9 * t)/t

So f'(t) = [t *ln9 e^(ln9 * t) - e^(ln9 * t) * 1]/t² (by quotient rule)

= [t ln9 *9^t - 9^t]/t²

= [9^t(tln9 - 1)]/t²

= 0 for stationary points

ie 9^t(tln9 - 1) = 0

ie tln9 - 1 = 0 (as 9^t ≠ 0)

so t = 1/ln9 ≈ 0.455

ie stationary point at (1/ln9, f(1/ln9))

f(1/ln9) = 9^(1/ln9)/(1/ln9)

= ln9*9^(1/ln9)

≈ 5.97 This is a local minimum and the least value of f(t) for t > 0

When t < 0, f'(t) < 0 so f(t) is monotonically decreasing

When 0
When t >1/ln9, f'(t) > 0 so f(t) is monotonically increasing

Answer 5

take f(t) as if it was y and then t as x
then it would look like
y=3^(2x)/x and then graph or something

Answer 6

You can use the computer calculator,it should help.

Answer 7

This is not math.

It is basic arithmetic.