Question

Find the right hand derivative of \[f\left( x \right) = \left[ x \right]\sin \pi x\] at \[x = n\], where \[n \in I\]

Answer 1

Hint: In this question, we need to determine the right hand derivative of the given function. For this, we will use the concept of the right hand limit at \[x = n + h\] such that h is infinitesimally small and tends to zero.

Complete step-by-step answer:
Given the function is \[f\left( x \right) = \left[ x \right]\sin \pi x\] where [.] is the greatest integer factor.
We have to find the right hand derivative of the function at \[x = n\] where n is an integer.
To determine the right hand limit of a function, we need to substitute the value of the parameter at the higher side of the defined point. Here, according to the question, we need to evaluate the limit at x=n so, we will substitute \[x = n + h\] where h is infinitesimally small and tends to zero.
We know the right hand derivative of a function is given by the formula \[\mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {c + h} \right) - f\left( c \right)}}{h}\], hence we can write the given function at right hand limit of the function as:
\[\Rightarrow f\left( {{x^ + }} \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left[ {n + h} \right]\sin \pi \left( {n + h} \right) - \left[ n \right]\sin \pi \left( n \right)}}{h}\]
Now since n is an integer which can be n=1, 2, 3, 4,… so, we can say \[\sin n\pi = 0\], hence we can further write the derivative as
\[\Rightarrow f\left( {{x^ + }} \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left[ {n + 1} \right]\sin \pi \left( {n + h} \right)}}{h}\]
We know \[\sin \left( {n\pi + x} \right) = {\left( { - 1} \right)^n}\sin x\]
Hence we can further write the derivative as
\[\Rightarrow f\left( {{x^ + }} \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left[ {n + 1} \right]{{\left( { - 1} \right)}^n}\sinh \pi }}{{h\pi }} \times \pi \]
Now since \[\mathop {\lim }\limits_{h \to 0} \dfrac{{\sinh \pi }}{{h\pi }} = 1\]
Hence we can further write the derivative
\[f\left( {{x^ + }} \right) = \pi \left[ {n + 1} \right]{\left( { - 1} \right)^n}\]
So the right hand derivative of the function \[f\left( x \right) = \left[ x \right]\sin \pi x\] at x=n\[ = \pi \left[ {n + 1} \right]{\left( { - 1} \right)^n}\]

Note: Right hand derivative of a function is \[\mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {c + h} \right) - f\left( c \right)}}{h}\]
Right hand derivative of a function f is defined as the right hand limit of a function. If the right hand derivative of a function exists then the function is said to be right hand differentiable.