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Question: Find the right hand derivative of \[f\left( x \right) = \left[ x \right]\sin \pi x\] at \[x = n\], w...

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

Explanation

Solution

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+hx = n + h such that h is infinitesimally small and tends to zero.

Complete step-by-step answer :
Given the function is f(x)=[x]sin⁡πxf\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=nx = 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+hx = 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 lim⁡h→0f(c+h)−f(c)h\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:
⇒f(x+)=lim⁡h→0[n+h]sin⁡π(n+h)−[n]sin⁡π(n)h\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⁡nπ=0\sin n\pi = 0, hence we can further write the derivative as
⇒f(x+)=lim⁡h→0[n+1]sin⁡π(n+h)h\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⁡(nπ+x)=(−1)nsin⁡x\sin \left( {n\pi + x} \right) = {\left( { - 1} \right)^n}\sin x
Hence we can further write the derivative as
⇒f(x+)=lim⁡h→0[n+1](−1)nsinh⁡πhπ×π\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 lim⁡h→0sinh⁡πhπ=1\mathop {\lim }\limits_{h \to 0} \dfrac{{\sinh \pi }}{{h\pi }} = 1
Hence we can further write the derivative
f(x+)=π[n+1](−1)nf\left( {{x^ + }} \right) = \pi \left[ {n + 1} \right]{\left( { - 1} \right)^n}
So the right hand derivative of the function f(x)=[x]sin⁡πxf\left( x \right) = \left[ x \right]\sin \pi x at x=n=π[n+1](−1)n = \pi \left[ {n + 1} \right]{\left( { - 1} \right)^n}

Note : Right hand derivative of a function is lim⁡h→0f(c+h)−f(c)h\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.