Question

Find the derivative.
y=3e^(-3/t)

Answer 1

I just did to you that kind of exercise a few moments ago
What is unclear tell me what you don't understand

Answer 2

Am I supposed to use the product rule?

Answer 3

You need to use the chain rule for this.

Answer 4

The product rule is for two functions multiplied together, not a constant and a function. The constant simply stays in the expression if it is a coefficient -like in this case.

Answer 5

once I rewrite it to look like this y=\[3e^{-3t ^{-1}}\] then what am I supposed to do?

Answer 6

derive it using chain rule.
D[f(g(x))]=f'(g(x))*g'(x)

F is the e function. -3(t^-1) is the g(x).

Answer 7

would this answer be y'=\[-3e^{-3t ^{-2} }\]

Answer 8

nah, try again.
f(x) is e^x.
g(x) is -3(t^-1)
Find f'(g(x))*g'(x). the -3 is just multiplied on at the end.
Start by deriving e^x. Evaluate that at g(x). Finally find g'(x). Multiply by g'(x). Multiply by -3. You will find the correct solution

Answer 9

Im sorry, I'm a person who learns better if I see it rather than explained

Answer 10

You have: \[3e ^{-3/t}\]
You need to use the chainrule. It states that:
\[\frac{ d }{ dx } f(g(x))=f'(g(x))*g'(x)\]

Answer 11

do you understand so far?

Answer 12

no really, I haven't dealt with the chain rule in a while since they gave us new rules for derivatives and integrals