How to use first principle to differenciate?? (1 Viewer)

[chrisk]

it does not count but everyone got zero - 33.33% because

the questions are crazy

You cant use anything else except for the first principle to differentiate these

a) tan x

b) Ln x

c) e^x

"c" is probably the easiest one, some of us could do it.

The math teacher is being gay and said "Not gonna give the answers
until end of the holiday"

help

 

[Trebla]

it does not count but everyone got zero - 33.33% because

the questions are crazy

You cant use anything else except for the first principle to differentiate these

a) tan x

b) Ln x

c) e^x

"c" is probably the easiest one, some of us could do it.

The math teacher is being gay and said "Not gonna give the answers
until end of the holiday"

help

LOL the questions are pretty nasty...I'll have a go at these...

Well the first principles for differentiation is:
f'(x) = lim_h-->0 [f(x + h) - f(x)] / h

a) f(x) = tan x
f(x + h) = (tan x + tan h) / (1 - tan x.tan h)
f(x + h) - f(x) = (tan x + tan h) / (1 - tan x.tan h) - tan x
= (tan x + tan h - tan x + tan²x.tan h) / (1 - tan x.tan h)
= (tan h + tan²x.tan h) / (1 - tan x.tan h)
= tan h(1 + tan²x) / (1 - tan x.tan h)

Now f'(x) = lim_h-->0 [f(x + h) - f(x)] / h
= lim_h-->0 [tan h(1 + tan²x) / h(1 - tan x.tan h)]
= lim_h-->0 [tan h / h].lim_h-->0 (1 + tan²x) / (1 - tan x.tan h)
but lim_h-->0 [tan h / h] = 1
and lim_h-->0 (1 + tan²x) / (1 - tan x.tan h) = (1 + tan²x)
.: f'(x) = 1 + tan²x
= sec²x (since sec²x - tan²x = 1)

b) f(x) = ln x
f(x + h) = ln (x + h)
f(x + h) - f(x) = ln (x + h) - ln x
= ln(1 + h/x)

Now f'(x) = lim_h-->0 [f(x + h) - f(x)] / h
= lim_h-->0 ln(1 + h/x) / h
= lim_h-->0 ln(1 + h/x)^1/h
Now let u = lim_h-->0 ln(1 + h/x)^1/h
=> e^u = lim_h-->0 (1 + h/x)^1/h
BUT by definition: e^x = lim_n-->∞ (1 + x/n)ⁿ
Now let n = 1/h and replace x with 1/x which gives:
e^1/x = lim_n-->∞ (1 + h/x)^1/h
.:e^u = lim_h-->0 (1 + h/x)^1/h
= e^1/x
.: u = 1/x
i.e. lim_h-->0 ln(1 + h/x)^1/h = 1/x
or f'(x) = 1/x


c) f(x) = e^x
f(x + h) = e^{x + h}
f(x + h) - f(x) = e^{x + h} - e^x
= e^x(e^h - 1)

Now f'(x) = lim_h-->0 [f(x + h) - f(x)] / h
= e^x lim_h-->0 (e^h - 1) / h

Need to prove that lim_h-->0 (e^h - 1) / h = 1
I'm not sure how to go about it (without resorting to L'Hopital's rule, Taylor series or assuming the gradient of y = e^x is 1 at x = 0) but here is a sloppy proof of it lol
We know that for some non-zero real number h:
e^h > h + 1 (this can be proved graphically or considering the minima of f(x) = e^x - x)
As h --> 0 from both negative or positive ends, both sides of the inequality converge to 1 (this is the "sloppy" bit as it assumes e^h and h + 1 converge at roughly the same rate but I guess it's a somewhat good approximation considering the Taylor series of e^h = 1 + h + h²/2! + h³/3! + .....)
.: (e^h - 1) / h --> 1 as h --> 0

Thus: f'(x) = e^x

Alternatively you could use the result from b)
Let y = e^x <=> x = ln y
dx/dy = 1/y
dy/dx = y
= e^x

 

[jet]

Yeah. Our teacher for 3 unit always told us L'hopitals rule for c).

 

[Trebla]

Yeah. Our teacher for 3 unit always told us L'hopitals rule for c).

The problem with using L'Hopital's rule for c) is that you use the result d(e^x)/dx = e^x to simplify the limit problem, which is the very result you are trying to prove!!!