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: Here in this question, we have to find the right hand derivative of given function.to solve this by using the formula of right hand derivative i.e., \[RHD = \mathop {\lim }\limits_{x \to {c^ + }} \dfrac{{f(x) - f(c)}}{{x - c}}\], on substituting the function of x and c and on further simplification we get the required solution.

Complete step by step answer:
The idea of a limit is a basis of all calculus. The limit of a function is defined as let \[f(x)\]be a function defined on an interval that contains\[x = a\]. Then we say that\[\mathop {\lim }\limits_{x \to a} f(x) = L\], if for every \[\varepsilon > 0\]there is some number \[\delta > 0\]such that \[\left| {f(x) - L} \right| < \varepsilon \]whenever\[0 < \left| {x - a} \right| < \delta \].
Consider the given function:
\[ \Rightarrow \,\,f\left( x \right) = \left[ x \right]\sin \pi x\] ---------(1)
Now, we haver to find the right hand derivative.
i.e., \[RHD = \mathop {\lim }\limits_{x \to {c^ + }} \dfrac{{f(x) - f(c)}}{{x - c}}\]-------(2)
Here to examine the differentiability, take some substitution for \[\mathop {\lim }\limits_{x \to {c^ + }} f(x)\] put \[x = n + h\] and change the limit as \[x \to {n^ + }\]by \[h \to 0\], then Equation (2) becomes
\[ \Rightarrow \,\,RHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(n + h) - f(n)}}{h}\]--------(3)
On substituting equation (2) in (3), then
\[ \Rightarrow \,\,RHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left[ {n + h} \right]\sin \pi \left( {n + h} \right) - \left[ n \right]\sin \pi n}}{h}\]
\[ \Rightarrow \,\,RHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {n + h} \right)\sin \left( {n\pi + h\pi } \right) - n\sin \pi n}}{h}\]
\[ \Rightarrow \,\,RHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {n + h} \right)\sin \left( {n\pi + h\pi } \right) - n\sin n\pi }}{h}\]
Now use the sine addition identity: \[sin\left( {A + B} \right) = sinA{\text{ }}cosB + cosA\,sinB\] applying in the above equation, then
\[ \Rightarrow \,\,RHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {n + h} \right)\left( {\sin \left( {n\pi } \right)\cos \left( {h\pi } \right) + \cos \left( {n\pi } \right)\sin \left( {h\pi } \right)} \right) - n\sin n\pi }}{h}\]
As we know the value of \[\cos n\pi = 1\] and \[\sin \left( {n\pi } \right) = 0\]. On substituting, we get
\[ \Rightarrow \,\,RHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {n + h} \right)\left( {0 \cdot 1 + \cdot 1\sin \left( {h\pi } \right)} \right) - n.0}}{h}\]
\[ \Rightarrow \,\,RHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {n + h} \right)\left( {0 + \sin \left( {h\pi } \right)} \right) - 0}}{h}\]
\[ \Rightarrow \,\,RHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{(n + h)(\sin (h\pi ))}}{h}\]
\[ \Rightarrow \,\,RHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{n\sin (h\pi ) + h\sin (h\pi )}}{h}\]
On simplification, we get
\[ \Rightarrow \,\,RHD = \mathop {\lim }\limits_{h \to 0} \left( {\dfrac{{n\sin (h\pi )}}{h} + \dfrac{{h\sin (h\pi )}}{h}} \right)\]
\[ \Rightarrow \,\,RHD = \mathop {\lim }\limits_{h \to 0} \dfrac{n}{h}\sin (h\pi ) + \mathop {\lim }\limits_{h \to 0} \dfrac{h}{h}\sin (h\pi )\]
\[ \Rightarrow \,\,RHD = \mathop {\lim }\limits_{h \to 0} \dfrac{n}{h}\sin (h\pi ) + \mathop {\lim }\limits_{h \to 0} \sin (h\pi )\]
On applying limit, we get the first term in the form of \[\dfrac{0}{0}\] form so we apply the L’Hospitals rule to the first term.
\[ \Rightarrow \,\,RHD = n\mathop {\lim }\limits_{h \to 0} h\,\cos (h\pi ) + \mathop {\lim }\limits_{h \to 0} \sin (h\pi )\]
Now when we apply the limit to the above function
\[ \Rightarrow \,\,RHD = n.0\,\cos (0.\pi ) + \sin (0.\pi )\]
Hence on simplification we get
\[ \Rightarrow \,\,RHD = 0\]
Hence, the right hand derivative of the given function is 0.

Note: The problem is related to the limits we have to find the right hand limit. We must know the formula \[RHD = \mathop {\lim }\limits_{x \to {c^ + }} \dfrac{{f(x) - f(c)}}{{x - c}}\] and we have to consider the given function and then we apply the limit to it. Since the given function is a trigonometry function we must know about the standard formulas of trigonometry.