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Theorems on Differentiability

  1. Addition of differentiable and non-differentiable function is always non-differentiable.
  2. Product of differentiable and non-differentiable function may be differentiable, e.g.,
     
    f(x) = x|x| is differentiable at x = 0
     
    f(x) = (x –1)|x –1| is differentiable at x = 1
     
    f(x) = (x –1)87780.png is differentiable at x = 1
     
    In general, f(x) = g(x)|g(x)| is differentiable at x = a when g(a) = 0. f(x) = x|x – 1| is non-differentiable at x = 1.
  3. If g(x) is differentiable function and f(x) = |g(x)| is non-differentiable at x = a, then g(a) = 0, e.g., |sin xis non-differentiable when sin x = 0 or x = nπn Z. But f(x) = |x3| is differentiable at x = 0, though f(0) = 0.
  4. If f(x) and g(x) both are non-differentiable at x = a then f(x) + g(x) may be differentiable at x = a, e.g.,
     
    f(x) = sin |x| – |x| = 87774.png
     
    g′(x) = 87767.png
     
    f′(0+) = 0 and f′(0) = 0

Differentiability using first definition of derivatives

Illustration
Discuss differentiability of 87761.png at x = 0
Solution
For continuity 87755.png
= 87749.png
 
Hence f(x) is continuous at x = 0.
 
Also f ′(0+) = 87743.png
= 87737.png
= 87731.png = 1
and f ′(0) = 87725.png
= 87719.png
 
Thus f(x) is differentiable at x = 0 and hence also continuous at x = 0.
 

Differentiability using theorems on differentiability

Illustration
Discuss differentiability of f(x) = |x| + |x – 1|.
 
Solution
f(x) = |x| + |x – 1|. f(x) is continuous everywhere as |x| and |x – 1| are continuous for all x. Also |x| and |x – 1| are non-differentiable at x = 0 and x = 1 respectively. Hence f(x) is non-differentiable at x = 0 and x = 1.
 
 
Illustration
Discuss the differentiability of f(x) = [x] + |1 – x|, x  (–1, 3), where [] is greatest integer function.
 
Solution
[x] is non-differentiable at x = 0, 1, 2 and |1 – x| is non-differentiable at x = 1. Thus f(x) is definitely non-differentiable at x = 0, 2. Moreover [x] is discontinuous at x = 1 whereas |1 – x| is continuous at x = 1. Thus f(x) is discontinuous and hence non-differentiable at x = 1.
 

Differentiability using graphs

Illustration
Discuss differentiability of
  1. f(x) = sin |x|
  2. f(x) = |loge|x||
  3. f(x) = sin–1(sin x)
  4. f(x) = maximum {2 sin x, 1 – cos x} ∀ x ∈ (0, π)
Solution
  1. f(x) = sin |x|
     
    87707.png
     
    Clearly from the graph f(x) is non-differentiable at x = 0.
  2. f(x) = |loge|x||
     
    87700.png
     
    Clearly from the graph f(x) is non-differentiable at x = 0, ±1.
  3. y = sin–1(sin x)
     
    87693.png
     
    Clearly from the graph f(x) is non-differentiable at x = 87687.png
  4. f(x) = maximum {2 sin x, 1 – cos x} can be plotted as
     
    87681.png
     
    Thus f(x) = maximum {2 sin x, 1 – cos x} is not differentiable, when 2 sin x = 1 – cos x.
     
    ⇒ 4 sin2 x = (1 – cos x)2
     
    ⇒ 4(1 + cos x) = (1 – cos x)
     
    ⇒ 4 + 4 cos x = 1 – cos x
     
    ⇒ cos x = –3/5
     
    ⇒ x = cos–1 (–3/5)
     
    Therefore, f(x) is not differentiable at x = π – cos–1 (3/5).

Differentiability by differentiation

Illustration
If f(x) = 87675.png, then find the values of b and c if f(x) is differentiable at x = 1
 
Solution
f(x) = 87669.png
⇒ f′(x) = 87663.png
 
f(x) is differentiable at x = 1. Then it must be continuous at x = 1. For which 87657.png
⇒ 1 + b + c = 1.
⇒ b + c = 0  ...(1)
 
Also f′(0+) = f′(0)
 
⇒ 87651.png
 
⇒ 2 + b = 1 ⇒ b = –1
 
⇒ c = 1 (from (1))
 




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