Differentiate This! (1 Viewer)

[Dreamerish*~]

1.1ⁿ with respect to n.

 

[香港!]

1.1ⁿ with respect to n.

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[香港!]

oh or is it 1.1^n?
i thought it was question 1.. different 1^n lol

anyway
y=1.1^n
ln y=n ln 1.1
n=(1\ln 1.1) ln y
dn\dy=(1\ln 1.1) (1\y)
dy\dn=flip it..

 

[icycloud]

1.1ⁿ with respect to n.

Let y = 1.1^n

ln(y) = ln(1.1^n)
ln(y) = n * ln(1.1)
1/y * dy/dn = ln(1.1)

dy/dn = ln(1.1) * y
= ln(1.1) * 1.1^n #

 

[Dreamerish*~]

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I'm afraid that is incorrect.

Oh, my bad. This is actually 2U. I feel like crap now.

Thanks Icycloud. Very smart.

 

[icycloud]

Haha hk beat me to it... ><

 

[香港!]

damn i tot she meant like.. "1." as in Question 1 LoLz

 

[Riviet]

damn i tot she meant like.. "1." as in Question 1 LoLz

1. 1 ---> Rofl!!

 

[KFunk]

1.1ⁿ with respect to n.

Another way you could do it is using the property (a)ⁿ = e^n.ln(a), if you do this then:

y = 1.1ⁿ = e^n.ln(1.1)

dy/dn = ln(1.1)e^n.ln(1.1) = ln(1.1)(1.1ⁿ) or = ln(1.1)y

 

[chin music]

y=1.1to the n
lny=ln(1.1 to the n)
lny=n*ln1.1 (plog rule or whatever other ppl call it)
n=1/ln1.1*(lny)
dn/dy=1/yln1.1
dy/dn = yln1.1
I reckon thats the best way of doing those

 

[ioniser]

hey do you people know that way with the e function trick to answer the question

to answer the above ,where you make the equation in terms of e it was done at the 4 unit day at that uni thingie,its really usefull if your in any trouble but i cant remember it

(and know it wasn't the solution as above)

 

[icycloud]

hey do you people know that way with the e function trick to answer the question

to answer the above ,where you make the equation in terms of e it was done at the 4 unit day at that uni thingie,its really usefull if your in any trouble but i cant remember it

(and know it wasn't the solution as above)

Let x = e^y

ln(x) = ln(e^y)
= yln(e)
=y

Therefore, x = e^(ln(x))

Is that what you wanted?