Homework Help Thread

[Eraddd]

THanks for showing to me man. Really appreciate it.

But could you find any nitpicks into how I did it? I don't understand how using the power rule, and using the chain rule screwed up =/

Man calculus is a bitch if you don't study. We started this a week and a half ago, and I just started now.

 

[mtr]

All right, I'll see what's going on.

Essentially, I divided the x and the e^(-kx) with the power rule, and with the chain rule, I could get the derivative of e^(-kx). I get...

But this is really it right here. Your mistake was approaching this as a power rule problem when in fact the structure doesn't lend itself to one. 'x' is a function, but your strategy interpreted it as a constant. Its like using a Hippowdon to deal with a Salamence that turns out to be a Choice Specs user! Basically, whenever you see what looks like 2 functions being multiplied by or divided by each other, you use the Product/Quotient Rule respectively. You will also see this theme in Integrals, where there are two basic strategies: Substitution and Parts, and you need to use the right one for the job.

 

[X-Act]

Sorry to bump this thread, but I need help with this question.

What is the derivative of y = xe^(-kx). The answer says it's [e^(-kx)][-kx+1] but I seem to get instead of -kx but -k^2x which is really frustrating me. Essentially, I divided the x and the e^(-kx) with the power rule, and with the chain rule, I could get the derivative of e^(-kx). I get

x d/dx e^(-kx)+ [e^(-kx)] d/dx x
which I believe is right.

But when I calculate it, I get:
d/dx e^(-kx) = [e^(-kx)][-kx] and then that's multiplied by x, so it equals [e^(-kx)][-(k^2)x]

Your mistake is here. d/dx e^(-kx) is not [e^(-kx)][-kx], but [e^(-kx)][-k], which you then multiply by x. So we have [e^(-kx)][-k] multiplied by x, which is equal to [e^(-kx)][-kx].

and the other side is just e^(-kx) as the derivative of x is 1.

This is correct.

So when I put these two things together, I get [e^(-kx)][-(k^2)x+1], which according to the answer key is wrong.

Anyone want to help?

Putting them together now, we get [e^(-kx)][-kx] + e^(-kx). You have e^(-kx) common, so the answer is [e^(-kx)][-kx + 1], exactly as your answer says.

So basically your mistake was that the derivative of e^(-kx) is equal to [e^(-kx)][-k], not [e^(-kx)][-kx]. It was just a slight mistake, then.

 

[Wishy]

Can anyone provide some chemistry help again? >:

We've started thermodynamics and enthalpy, and we have to complete this chart but I have no clue what to do

We're given the net ionic equation:

NaOH -> Na+ + OH-

What is the 'heat released by Process (J)'? we're told the density of all solutions should be assumed to be = to the density of water, and the heat capacity is 4.18 J/g x(times) degrees (in Celsius)

???

 

[Eraddd]

Its like using a Hippowdon to deal with a Salamence that turns out to be a Choice Specs user!

Loved the Pokemon Analogy.

@X-act: Thanks for the help. I really appreciate it. Finally know the source of my error. I'll keep my head up next time thanks.

 

[sbc]

Let's try using the division rule, though I foresee this getting messy anyways.

xe^(-kx) = x/(e^kx)

So immediately we have f(x) as 'x' and g(x) as 'e^kx'

(f/g)' = (gf' - fg')/(g^2)

So we have (e^kx - kx*e^-kx) / (e^kx)^2

Factor out e^kx from the top and from the bottom to get [(e^kx)(1-kx)] / [(e^kx)(e^kx)]

Now divide the numerator and denominator by e^kx.

So we get (1-kx)/(e^kx).

This is the same thing that your book says, just written a different way via basic exponent rules.

Would've been easier in this case to use the product rule. If you made u=x and v=e^-kx

dy/dx = v (du/dx) + u (dv/dx)
dy/dx = 1 (e^-kx) + x (-ke^-kx)
dy/dx = e^-kx - xk(e^-kx)
then factorise to get the answer.
dy/dx = (1-kx)e^-kx

You could use quotient as you did, but it's overly long and is more likely to lead to mistakes when since you have to get the top line the right way round and you also need to cancel out your answer which is just another line of working.

 

[Bologo]

Can anyone provide some chemistry help again? >:

We've started thermodynamics and enthalpy, and we have to complete this chart but I have no clue what to do

We're given the net ionic equation:

NaOH -> Na+ + OH-

What is the 'heat released by Process (J)'? we're told the density of all solutions should be assumed to be = to the density of water, and the heat capacity is 4.18 J/g x(times) degrees (in Celsius)

???

We need a little more information. Does it tell you any temperature changes, or mass, or even volume (which will basically be the same as mass due to the equal density with water)?

 

[mtr]

Would've been easier in this case to use the product rule. If you made u=x and v=e^-kx

dy/dx = v (du/dx) + u (dv/dx)
dy/dx = 1 (e^-kx) + x (-ke^-kx)
dy/dx = e^-kx - xk(e^-kx)
then factorise to get the answer.
dy/dx = (1-kx)e^-kx

You could use quotient as you did, but it's overly long and is more likely to lead to mistakes when since you have to get the top line the right way round and you also need to cancel out your answer which is just another line of working.

Well, I always screw up with integer signs, so I prefer it if they are out of the picture. You can't go wrong either way, though.