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Reply 1
Do you mean y = 2e^(2x+1)?
Look for the thing you can differentiate. You can easily differentiate 2e^(something), so set u = 2x+1.
Look for the thing you can differentiate. You can easily differentiate 2e^(something), so set u = 2x+1.
generalebriety
Do you mean y = 2e^(2x+1)?
Look for the thing you can differentiate. You can easily differentiate 2e^(something), so set u = 2x+1.
Look for the thing you can differentiate. You can easily differentiate 2e^(something), so set u = 2x+1.
not sure, i think it is not a power, take a look at the site i got it from:
http://www.s-cool.co.uk/topic_quicklearn.asp?loc=ql&topic_id=8&quicklearn_id=4&subject_id=1&ebt=171&ebn=&ebs=&ebl=&elc=13
question 3, and at the bottom they give the answer to be dy/dx = 4e(2x + 1) , so how was this answer reached?
Reply 3
Hmm, that is strange. I think it's meant to be a power - apply the method I said and you should get the answer 4e^(2x+1). (If you differentiate what they wanted you to, you get 4e, and there's no way you can get 4e(2x+1), so I can only assume they've messed the superscripts up. 
)
)generalebriety
Hmm, that is strange. I think it's meant to be a power - apply the method I said and you should get the answer 4e^(2x+1).
yep, that what i got too. and you're right, the website is at fault as it has not written it with powers, otherwise our answer matches with the sites.
thanks for your prompt help generalebriety, +ive rep comming your way.
Reply 5
No problem. 
i'm actually learning c3 diff. from that site, i'm having trouble understanding how y=e^3x differentiates to dy/dx = 3 e^3x.
i know that the function is identical to the gradient function, but where does the coefficient 3 come from.
i diff. y=e^3x by first principal but only got dy/dx = e^3x ...(i think!), what about the 3 that the site gives infront of e?
i know that the function is identical to the gradient function, but where does the coefficient 3 come from.
i diff. y=e^3x by first principal but only got dy/dx = e^3x ...(i think!), what about the 3 that the site gives infront of e?
arslan
i'm actually learning c3 diff. from that site, i'm having trouble understanding how y=e^3x differentiates to dy/dx = 3 e^3x.
i know that the function is identical to the gradient function, but where does the coefficient 3 come from.
i diff. y=e^3x by first principal but only got dy/dx = e^3x ...(i think!), what about the 3 that the site gives infront of e?
i know that the function is identical to the gradient function, but where does the coefficient 3 come from.
i diff. y=e^3x by first principal but only got dy/dx = e^3x ...(i think!), what about the 3 that the site gives infront of e?
differentiate the power and that is the coefficent.
Reply 8
General rule:
y = e^f(x), dy/dx = f'(x)e^f(x)
That should be in whichever book your using.
y = e^f(x), dy/dx = f'(x)e^f(x)
That should be in whichever book your using.
Reply 9
arslan
i'm actually learning c3 diff. from that site, i'm having trouble understanding how y=e^3x differentiates to dy/dx = 3 e^3x.
i know that the function is identical to the gradient function, but where does the coefficient 3 come from.
i diff. y=e^3x by first principal but only got dy/dx = e^3x ...(i think!), what about the 3 that the site gives infront of e?
i know that the function is identical to the gradient function, but where does the coefficient 3 come from.
i diff. y=e^3x by first principal but only got dy/dx = e^3x ...(i think!), what about the 3 that the site gives infront of e?
y = e^3x
You can differentiate e^(blah), but not e^3x, so let u = 3x.
y = e^u
dy/du = e^u
du/dx = 3
dy/dx = dy/du * du/dx = 3e^u
But u = 3x
dy/dx = 3e^3x.
In general, dy/dx = dy/du * du/dx.
Reply 10
arslan
i diff. y=e^3x by first principal but only got dy/dx = e^3x ...(i think!), what about the 3 that the site gives infront of e?
Ok, and by first principles:
y = e^3x.
P1(x,e3x)
P2(x+δx,e3(x+δx))
m = [e3(x+δx) - e3x] / [x + δx - x]
= [e3(x+δx) - e3x] / δx
= e3x(e3δx - 1) / δx.
lim(δx->0) m = e3x [lim(δx->0) (e3δx - 1) / δx].
Definition of derivative:
[f(z) - f(a)] / (z-a) -> f'(a) as z -> a.
(Here, I am taking f(z) = e^z, z = δx, a = 0.)
lim(δx->0) (e3δx - 1) / δx = 3 * lim(δx->0) (e3δx - e0) / (3δx - 0) = 3e^0 = 3.
Therefore,
lim(δx->0) m = e3x * 3
dy/dx = 3e^3x.
Phew... I think that's right, someone check that for me please?
Reply 12
Clever, but implicit differentiation isn't on until C4, so I'm guessing the OP won't have done this yet. Anyway, chain rule is quicker in your head... well, in my head. 
Edit: oh, and on your second step you used the chain rule too.
Anyway, chain rule is quicker in your head... well, in my head. Edit: oh, and on your second step you used the chain rule too.
Reply 14
Original post by CalculusMan
General rule:
y = e^f(x), dy/dx = f'(x)e^f(x)
That should be in whichever book your using.
y = e^f(x), dy/dx = f'(x)e^f(x)
That should be in whichever book your using.
This was really helpful, thank you for taking the time to post calculus man
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