Question

f(t)=7He^t+6L(ln(t))

where H and L are constants.
What is the derivative?

Answer 1

@mathmale

Answer 2

Note carefully that you have 2 capital letters, H and L, which represent constants. Treat them in the same manner as you would treat numerals such as 29, 13, 2, etc.

f(t)=7He^t+6L(ln(t)) in simplest form would look like f(t)=e^t - ln t.

Find the derivative of this new f(t) now, please, for practice.

Answer 3

f '(t) = ?

Answer 4

f'(t)=e^t+1/t

Answer 5

Perfect.

Now, differentiate 10e^t-4.5ln t

Answer 6

where did you get 10 and -4.5

Answer 7

Strictly examples. Just like those mysterious L and H.

Answer 8

Intentionally exposing you to new situations so that they'll be familiar in the future.

Answer 9

10e^t-(9/2t)

Answer 10

Your 10e^t is fine.
But your (9/2t) is ambiguous. How would you fix that?

(d/dt) [-4.5 ln t] = ?? Hint: Write (-4.5) first. Then, differentiate ln t. Lastly, multiply these 2 results together.

Answer 11

-4.5/t

Answer 12

Much better. Clear, accurate. Thx again for your perseverance.

Answer 13

:D

Answer 14

Differentiate: f(t)=7He^t+6L(ln(t))

Answer 15

7e^(t)+(6/t)?

Answer 16

What hap to the 'H?'

Answer 17

don't constants come out of the equation?

Answer 18

Not in cases like this: That H and that L are constant coefficients.
The H stays with the first term and the L stays wtih the second: f(t)=7He^t+6L(ln(t))

Try again. Hint: 7H is a constant coeff.

Answer 19

(7He^t+(6L/t))

Answer 20

Examples: (d/dt)[Ht]=H
but (d/dt) [H] = 0. Why the difference?

Yes, you're right. Be sure you label your result. What is the name of the original function you were supposed to differentiate?

Answer 21

f'(t)

Answer 22

Right. So now you have a completely correct result, with label, with H and L treated like the constant coeffs they are. Very good.

Answer 23

You have f '(t) = (7He^t+(6L/t))

I would first omit the first and last parentheses: 7He^t (no left parenthesis) AND (6L)(1/t) (Your parentheses in (6L/t) are in the wrong place; this expression should look like (6L)(1/t) or simply (6L)/t or even 6L/t.

Answer 24

Gotta get off the 'Net. Looking forward to working with you again. ;)