Test: Continuity And Differentiability - 2

L'Hopital's rules says that the  
lim x→a f(x)/g(x)
⇒ f'(a)/g'(a)
Using this, we get  
lim x→0 (1−cosx)/x²
⇒ − sin0/2(0)
Yet as the denominator is 0, this is impossible. So we do a second limit:
lim(x→0) sinx/2x
⇒ cos0/2 = 1/2 = 0.5
So, in total lim x→0 (1−cosx)/x²
⇒ lim x→0 sinx/2x
⇒ cosx/2
⇒ cos0/2= 1/2

As all the three remaining statements are true for the given function.

f(g(x)) = x 
f'(g(x)) g'(x) = 1 
put x = a
f'(b) g'(a) = 1
2 f'(b) = 1
f'(b) = 1/2

The graph of f(x) = |x|
As observed from the graph, f(x) = |x| is continuous at x = 0.
As this curve is pointed at x = 0, f(x) is not derivable at x = 0.

Given x^py^q = (x+y)^p+qTaking log on both sides we get:
_plogx+_qlogy = (p+q) log (x+y).  Differentiating both sides w.r.t. x we get , 

y = ae^mx + be^-mx ⇒ y1 = ame^mx + (-m)be^-mx ⇒y2

= am²e^mx _{+ (m}²)be^-mx ⇒y2 = m² (ae^mx + be^-mx) ⇒ y2 = m²y

y  = a sin mx+b cos mx ⇒ y₁ = am cos mx − bm sin mx
⇒ y₂ = −am²sin mx−bm²cosmx ⇒ y₂ = −m²(a sin mx+b cos mx)

 (by using L’Hospital Rule)