1. ## Derivation

Hey Guys Hope you are Fine!

i have a problem and i need to know the derivation of this:
lim : 1-sin^2x/ x-pi/2
x-->pi/2

For me we should use Hospital so:
Lim: -2sin2x/ 1-pi/2 ? right? and then?
x--->pi/2

and my problem :

f (x) =x + Radical (1 + x ^ 2)
A) proove that x + (Radical) 1 + x ^ 2> 0
B) Calculate derivative of the function f ' [i did this],f and proove that radical (1 + x ^ 2) f' (x) = f (x)
Deduisez in that reel for all x, f (x)> 0
C) proove that
(1 + x ^ 2) f '(x) + xf' (x) = f (x)

Thanks a lot

2. Any help? i am going to school is 15 mins

3. Originally Posted by iceman1
Hey Guys Hope you are Fine!

i have a problem and i need to know the derivation of this:
lim : 1-sin^2x/ x-pi/2
x-->pi/2

For me we should use Hospital so:
Lim: -2sin2x/ 1-pi/2 ? right? and then?
x--->pi/2
now plug in $\displaystyle x = \frac {\pi}2$

Originally Posted by iceman1
f (x) =x + Radical (1 + x ^ 2)
A) proove that x + (Radical) 1 + x ^ 2> 0
The result is obviously true for $\displaystyle x \ge 0$, since $\displaystyle \sqrt{1 + x^2} > 0$ for all $\displaystyle x$.

Thus, we consider if $\displaystyle x < 0$

again, since $\displaystyle \sqrt{1 + x^2} > 0$ for all $\displaystyle x$, we need not worry about it making our function negative. thus it suffices to show that $\displaystyle \sqrt{1 + x^2} > x$ for $\displaystyle x < 0$. (since that would show that no matter how negative $\displaystyle x$ gets, it will always be overwhelmed by the more positive $\displaystyle \sqrt{1 + x^2}$, and thus $\displaystyle f(x) = x + \sqrt{1 + x^2} > 0$ for all $\displaystyle x$).

Now, consider the function $\displaystyle g(x) = \sqrt{1 + x^2} - x$

we claim that $\displaystyle g(x) > 0$. For consider otherwise that $\displaystyle g(x) \le 0$

if $\displaystyle g(x) = 0$, then $\displaystyle \sqrt{1 + x^2} - x = 0 \implies \sqrt{1 + x^2} = x \implies 1 + x^2 = x^2$, which clearly makes no sense

likewise, if $\displaystyle g(x) < 0$, then $\displaystyle \sqrt{1 + x^2} - x < 0 \implies 1 + x^2 < x^2$, again makes no sense. thus, $\displaystyle g(x) > 0$

but that means that $\displaystyle \sqrt{ 1 + x^2} - x > 0$ which means that $\displaystyle \sqrt{1 + x^2} > x$. thus, we have proven that $\displaystyle f(x) > 0$ for all $\displaystyle x$

QED

B) Calculate derivative of the function f ' [i did this],f and proove that radical (1 + x ^ 2) f' (x) = f (x)
Deduisez in that reel for all x, f (x)> 0
C) proove that
(1 + x ^ 2) f '(x) + xf' (x) = f (x)

Thanks a lot
if you found f'(x) already, what is your problem here? just plug it in and solve. plug in f'(x) into the left side of the equations in B) and C) and show that when you simplify, you get the right hand side, namely, f(x), the original function