# Theorems on Differentiability

- Addition of differentiable and non-differentiable function is always non-differentiable.
- Product of differentiable and non-differentiable function may be differentiable, e.g.,
*f*(*x*) =*x*|*x*| is differentiable at*x*= 0*f*(*x*) = (*x*â€“1)|*x*â€“1| is differentiable at*x*= 1*f*(*x*) = (*x*â€“1) is differentiable at*x*= 1*f*(*x*) =*g*(*x*)|*g*(*x*)| is differentiable at*x*=*a*when*g*(*a*) = 0.*f*(*x*) =*x*|*x*â€“ 1| is non-differentiable at*x*= 1. - If
*g*(*x*) is differentiable function and*f*(*x*) = |*g*(*x*)| is non-differentiable at*x*=*a*, then*g*(*a*) = 0, e.g., |sin*x*| is non-differentiable when sin*x*= 0 or*x*=*n**Ï€*,*n*âˆˆ*Z*. But*f*(*x*) = |*x*^{3}| is differentiable at*x*= 0, though*f*(0) = 0. - 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*| â€“ |*x*| =*g*â€²(*x*) =*f*â€²(0^{+}) = 0 and*f*â€²(0^{â€“}) = 0

# Differentiability using first definition of derivatives

**Illustration**

Discuss differentiability of at

*x*= 0Solution

For continuity

=

Hence

*f*(*x*) is continuous at*x*= 0.Also

*f*â€²(0^{+}) ==

= = 1

and

*f*â€²(0^{â€“}) ==

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*â€“ 1|.Solution

*f*(

*x*) = |

*x*| + |

*x*â€“ 1|.

*f*(

*x*) is continuous everywhere as |

*x*| and |

*x*â€“ 1| are continuous for all

*x*. Also |

*x*| and |

*x*â€“ 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 â€“*x*|,*x*âˆˆ (â€“1, 3), where [â‹…] is greatest integer function.Solution

[

*x*] is non-differentiable at*x*= 0, 1, 2 and |1 â€“*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 â€“*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

*f*(*x*) = sin |*x*|*f*(*x*) = |log|_{e}*x*||*f*(*x*) = sin^{â€“1}(sin*x*)*f*(*x*) = maximum {2 sin*x*, 1 â€“ cos*x*} âˆ€*x*âˆˆ (0,*Ï€*)

Solution

|*f*(*x*) = sin|*x**f*(*x*) is non-differentiable at*x*= 0.|*f*(*x*) =**log**|_{e}||*x**f*(*x*) is non-differentiable at*x*= 0, Â±1.*y*= sin^{â€“1}(sin*x*)*f*(*x*) is non-differentiable at*x*=*f*(*x*) = maximum {2 sin*x*, 1 â€“ cos*x*} can be plotted as*f*(*x*) = maximum {2 sin*x*, 1 â€“ cos*x*} is not differentiable, when 2 sin*x*= 1 â€“ cos*x.*^{2}*x*= (1 â€“ cos*x*)^{2}*x*) = (1 â€“ cos*x*)*x*= 1 â€“ cos*x**x*= â€“*x*= cos^{â€“1}(â€“3/5)*f*(*x*) is not differentiable at*x*=*Ï€*â€“ cos^{â€“1}(3/5).

# Differentiability by differentiation

**Illustration**

If

*f*(*x*) = , then find the values of*b*and*c*if*f*(*x*) is differentiable at*x*= 1Solution

*f*(

*x*) =

â‡’

*f*â€²(*x*) =*f*(

*x*) is differentiable at

*x*= 1. Then it must be continuous at

*x*= 1. For which

â‡’ 1 +

*b*+*c*= 1.â‡’

*b*+*c*= 0 ...(1)Also

*f*â€²(0^{+}) =*f*â€²(0^{â€“})â‡’

â‡’ 2 +

*b*= 1 â‡’*b*= â€“1â‡’

*c*= 1 (from (1))