Question

The value of derivative of \[\left| {x - 1} \right| + \left| {x - 3} \right|\] at \[x = 2\] is:
\[\
  \left( A \right)\,\,\,\,2 \\
  \left( B \right)\,\,\,\,1 \\
  \left( C \right)\,\,\,\,0 \\
  \left( D \right)\,\,\,\, - 2 \\
\ \]

Answer 1

Hint: The derivative of a function of real variable measure the sensitivity to change of a quantity which is determined by another quantity .The derivative of f(x) with respect to x is the function\[f'\left( x \right)\] and is defined as , \[f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\].

Complete step-by-step answer:
The objective of the problem is to find the value of derivative of \[\left| {x - 1} \right| + \left| {x - 3} \right|\] at \[x = 2\]
We have to find the value of the derivative of the given function at two.
Let the given function be \[f\left( x \right) = \left| {x - 1} \right| + \left| {x - 3} \right|\]
To find the derivative we use the formula \[f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\]
It is given in the problem to find the value at x=2 the above derivative formula becomes \[f'\left( 2 \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {2 + h} \right) - f\left( x \right)}}{h}........\left( 1 \right)\]
First we find the value of \[f\left( 2 \right)\].
It is nothing but substituting two in the place of x in f(x).
That is \[\
  f\left( 2 \right) = \left| {2 - 1} \right| + \left| {2 - 3} \right| \\
  \,\,\,\,\,\,\,\,\,\,\, = \left| 1 \right| + \left| { - 1} \right| \\
\ \]
The minus value in modulus becomes positive.
Now \[f\left( 2 \right) = 1 + 1 = 2\].
Therefore the value of \[f\left( 2 \right)\] is two .
Now we have to find the value of \[f\left( {2 + h} \right)\].that is substituting 2+h in the place of x in f(x).
\[\
  f\left( {2 + h} \right) = \left| {2 + h - 1} \right| + \left| {2 + h - 3} \right| \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left| {1 + h} \right| + \left| {h - 1} \right| \\
\ \]
Now substitute the values of \[f\left( 2 \right)\] and \[f\left( {2 + h} \right)\] in equation (1), we get
\[f'\left( 2 \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left[ {\left| {1 + h} \right| + \left| {h - 1} \right|} \right] - 2}}{h}\]
Since the h is almost equal to zero the modulus values of 1+h is positive and the modulus value of h-1 is negative.
\[\
  f'\left( 2 \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{1 + h - h + 1 - 2}}{h} \\
  \,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{h \to 0} \dfrac{{1 + h - h + 1 - 2}}{h} \\
  \,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{h \to 0} \dfrac{0}{h} = 0 \\
\ \]
Thus, the derivative of a given function at x=2 is zero.
Therefore , option C is correct.

Note: We can use the same formula to find derivatives of any function . All the standard formulas of the functions are derived from the above formula only. To solve these types of problems one should avoid common mistakes by making proper use of functional notation and careful use of basic algebra.