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Solved Examples on Differentiability Illustration 1: Let [.] denotes the greatest integer function and f(x) = [tan2x], then does the limit exist or is the function differentiable or continuous at 0? (1999) Solution: Given f(x) = [tan2x] Now, -45°< x < 45° tan(-45°)< tanx < tan45° -tan 45°< tan x < tan 45° -1< tan x <1 So, 0 <tan2x < 1 [tan2x] = 0 So, f(x) is zero for all values of x form x = -45° to 45°. Hence, f is continuous at x =0 and f is also differentiable at 0 and has a value zero. Illustration 2: A function is defined as follows: f(x)= x3 , x2< 1 x , x2≥ 1 Discuss the differentiability of the function at x=1. Solution:We have R.H.D. = Rf'(1) = limh→0 (f(1-h)-f(1))/h = limh→0 (1+h-1)/h = 1 and L.H.D. = Lf'(1)= limh→0 (f(1-h)-f(1))/(-h) = limh→0 ((1-h)3-1)/(-h) = limh→0 (3-3h+h2) = 3 ?Rf'(1)≠ Lf'(1)⇒ f(x) is not differentiable at x=1. Illustration 3:If y = (sin-1x)2 + k sin-1x, show that (1-x2) (d2 y)/dx2 - x dy/dx = 2 Solution: Here y = (sin-1x)2 + k sin-1x. Differentiating both sides with respect to x, we have dy/dx = 2(sin-1 x)/√(1-x2 ) + k/√(1-x2 ) ⇒(1-x2 ) (dy/dx)2 = 4y + k2 Differentiating this with respect to x, we get (1-x2) 2 dy/dx.(d2 y)/(dx2 ) - 2x (dy/dx)2 = 4(dy/dx) ⇒(1-x2 ) ( d2 y)/dx2 -x dy/dx = 2
Illustration 1: Let [.] denotes the greatest integer function and f(x) = [tan2x], then does the limit exist or is the function differentiable or continuous at 0? (1999)
Solution: Given f(x) = [tan2x]
Now, -45°< x < 45°
tan(-45°)< tanx < tan45°
-tan 45°< tan x < tan 45°
-1< tan x <1
So, 0 <tan2x < 1
[tan2x] = 0
So, f(x) is zero for all values of x form x = -45° to 45°.
Hence, f is continuous at x =0 and f is also differentiable at 0 and has a value zero.
Illustration 2: A function is defined as follows:
f(x)= x3 , x2< 1
x , x2≥ 1
Discuss the differentiability of the function at x=1.
Solution:We have R.H.D. = Rf'(1)
= limh→0 (f(1-h)-f(1))/h
= limh→0 (1+h-1)/h = 1
and L.H.D. = Lf'(1)= limh→0 (f(1-h)-f(1))/(-h)
= limh→0 ((1-h)3-1)/(-h)
= limh→0 (3-3h+h2) = 3
?Rf'(1)≠ Lf'(1)⇒ f(x) is not differentiable at x=1.
Illustration 3:If y = (sin-1x)2 + k sin-1x, show that (1-x2) (d2 y)/dx2 - x dy/dx = 2
Solution: Here y = (sin-1x)2 + k sin-1x.
Differentiating both sides with respect to x, we have
dy/dx = 2(sin-1 x)/√(1-x2 ) + k/√(1-x2 )
⇒(1-x2 ) (dy/dx)2 = 4y + k2
Differentiating this with respect to x, we get
(1-x2) 2 dy/dx.(d2 y)/(dx2 ) - 2x (dy/dx)2 = 4(dy/dx)
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