How Do You Apply the Product Rule to Differentiate 19te^-4t?

[jpd5184]

Homework Statement

find the derivative of x=19te^-4t

Homework Equations

chain rule and the product rule

The Attempt at a Solution

i split the equation into 19t and e^-4t.

the derivative of 19t is 19.
the derivative of e^-4t is ? (is it 19/te^4)

 

[Mark44]

Homework Statement

find the derivative of x=19te^-4t

Homework Equations

chain rule and the product rule

The Attempt at a Solution

i split the equation into 19t and e^-4t.

the derivative of 19t is 19.
the derivative of e^-4t is ? (is it 19/te^4)

There's also the constant multiply rule: d/dt(k*f(t)) = k * d/dt(f(t)).

This means that d/dt(19te^(-4t)) = 19*d/dt(te^(-4t))

Now use the product rule on te^(-4t). You will need the chain rule when you differentiate e^-(4t).

Note that 19te^(-4t) is NOT an equation.

 

[jpd5184]

is there anyway somebody could give me the answer with all the steps because i just don't get it and i have tried it a million times and i really need the answer for homework. thanks.

 

[HallsofIvy]

The question you asked was really about the derivative of e^(-4t). Notice that this is NOT of the form t^n for any number n- t is the exponent, not the base.

The derivative of e^u with respect to u is just e^u again. Using the chain rule, if u is a function of t, de^u/dt= de^u/du du/dt= e^u du/dt. In particular, if u= -4t, then du/dt= -4 so the derivative of e^(-4t) is -4e^(-4t).

Now, try differentiating t^2 e^(-4t) using the product rule.

 

[Mark44]

is there anyway somebody could give me the answer with all the steps because i just don't get it and i have tried it a million times and i really need the answer for homework. thanks.

You're relatively new here, so probably don't know how it works. We will help you do the work, but the policy is that we don't just give you the answer, even if you have tried it a million times.

 

[gerben]

You just seem to have trouble understanding the chain rule, try to see the function f = e^-4t as one function inside another.
First there is a simple function of t, let's call it u:
u = -4t
then you should see that e^-4t is in fact e to the power "this function u":
e^u.

Now the chain rule says that the derivative of e^-4t, is the product of the derivatives of these two functions. It is the product of the derivative of -4t and the derivative of e^u.

So
what is the derivative of -4t? say it is A
what is the derivative of e^u? say it is B
Then
the derivative of e^-4t is: A*B

 

[Dickfore]

hint: product rule

$$(u v)' = u' v + u v'$$

 

[jpd5184]

anybody please i really need the answer.

i know how to use the product rule. just tell me what two functions to use for the product rule.

plus I am not sure the derivative of e^-4t

 

[Dick]

i) you already suggested in your first post to split it into u=19t and v=e^(-4t). That's fine. ii) you just said you knew how to use the product rule. iii) gerben and hallsofivy both explained how to use the chain rule on e^(-4t) and halls even told you the answer. And iv) Mark44 said we won't tell you the answer because of forum rules. How much more help do you need?

 

[HallsofIvy]

anybody please i really need the answer.

No, you don't "really need the answer"- you really need to know how to get the answer and you really need to know how to think about such problems. That's why you need to work it out yourself as much as possible.

i know how to use the product rule. just tell me what two functions to use for the product rule.

Well, ignoring the constant, 19, what two functions are multiplied here?

plus I am not sure the derivative of e^-4t

Which proves you don't really need the answer! I told you what the derivative was back in the third response. Obviously "giving" you the answer doesn't help.