trial question (1 Viewer)

[fashionista]

heeeeey i was doin last yrs trial from my school and i cant do this question and help would be appreciated sooo much. its even more frustrating cuz its a one mark question.

here it is:
(i)use first principles to show that the derivative of f(x)=2^x is f'(x)=c.2^x, where c is some constant.
A: so i did this bit and c= [(2^h)-1]/h
this bit is the bit i cant do
(ii) estimate the value of c correct to 2 d.p
PLEASE HELP!!!!!
luv me.

 

[coca cola]

have you tried to differentiate f(x) using normal method then equating c?

covert 2^x to a base of e, then differentiate it.

 

[coca cola]

c should equals to 1/ln2 if i didn't make a mistake with my logs.

then you just type that in the calculator by using logs again... so its 1.44

umm i should of checked with my log rules... love you

 

[CrashOveride]

f(x) = 2^x Then f'(x) = ln(2). 2^x
So c = ln(2)

Of course, this isn't first principles...having posted such a question i dread i will actually have to go and re-learn such material.

For the estimation part, i'm thinking perhaps go with Newton's method.

 

[coca cola]

f(x) = 2^x Then f'(x) = ln(2). 2^x
So c = ln(2)

Of course, this isn't first principles...having posted such a question i dread i will actually have to go and re-learn such material.

For the estimation part, i'm thinking perhaps go with Newton's method.

umm i don't think you understood the question mate. the first principle is already done, all you have to do is equate the c with normal differential solution. your answer seems wrong by inspection because e > 2 , hence c must be bigger than 1. ln2 is 0.6...

first principle is simple just substitute the value in and factorise to get 2^x * [(2^h)-1]/h

 

[CrashOveride]

The normal differential solution for the constant part is ln2

 

[coca cola]

ok, i must be wrong then, seems that i got my logics wrong with the inspection part.

if you want to do that question a mechanical/iterative way like the newton-raphson method, you can, but just by substituting values very very close to 0 into h: [(2^h)-1]/h.

about 3 or 4 d.p should give you the required appromate 2 d.p solution.

i tried with 0.0001, it gave me 0.69... which equals to CrashOveride's ln2.

 

[fashionista]

thanks soo much for ur help!!!
thank u thank u

 

[fashionista]

ohh ohh i have another question *embarrassed face*, this one isn't as trivial as the other one.

Here it is!! (oh..i can't do part (ii)):

 

[Supra]

AP=xcot60
BP=xcot30
using BP²=AP²+1
u find x by:
(xcot30)²=(xcot60)²+1
(xcot30)²-(xcot60)²=1
x²{cot²30-cot²60}=1
x²=1/{cot²30-cot²60}
x=SqRoot [1/{cot²30-cot²60}]
x=0.612372435km
x=610m --> i dont know if this is correct, the helicopter wud b very high up

 

[Estel]

Quote: i dont know if this is correct, the helicopter wud b very high up

i.)
AP = xtan30 = x/rt3
BP = xtan60 = xrt3

ii.)
In APB
(x/rt3)^2 + 1 = (xrt3)^2
Hence (8x^2)/3 = 1
x = rt[3/8] in km
= 610 m

you are correct supra

 

[fashionista]

YAY!!! thanku!!!!!!!!

 

[mr EaZy]

AP=xcot60
x=0.612372435km
x=610m --> i dont know if this is correct, the helicopter wud b very high up

Thats nothing for a chopper! or is it normal...