Question

If a is not a multiple of $\pi $, then show that the function$g\left( x \right)=\csc x$ is differentiable at a and $g'\left( a \right)=-\csc a\cot a$. In general, $g'\left( x \right)=-\csc x\cot x$ for all $x$$\ne $$n\pi $, where $n\in Z$.

Answer 1

Hint: Use the fundamental definition for proving any function to be differentiable or not which is given as. If any function $f\left( x \right)$is differentiable at point ‘$c$’ then LHD and RHD should be equal which are given by relation

Complete step-by-step answer:
LHD $=\underset{\lambda \to {{c}^{-}}}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( c \right)}{x-c}$ and RHD $=\underset{x\to {{c}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( c \right)}{x-c}$

As we know that any function $f\left( x \right)$is differentiable at any point c, if it’s Left hand derivative (LHD) and Right hand derivative (RHD) are equal to each other and equal to $f'\left( c \right)$ as well.
 LHD and RHD of function $f\left( x \right)$ at any point ‘c’ can be given as
LHD $=\underset{\lambda \to {{c}^{-}}}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( c \right)}{x-c}$ ………………………………………………(i)
RHD $=\underset{x\to {{c}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( c \right)}{x-c}$ ………………………………………………(ii)
Hence, any function $f\left( x \right)$is differentiable at point ‘c’ if
LHD $=$RHD $=f'\left( c \right)$ ……………………………………………………(iii)
Now coming to the question, we have a function$g\left( x \right)=\csc x$ where we need to prove it is differentiable at $x=a$where $a$is not multiple of $\pi $i.e. $a$$\ne $ $n\pi $.
And, we have given in question that $g'\left( x \right)=-\csc x\cot x$, so $g'\left( a \right)=-\csc a\cot a$
For all $x$$\ne $$n\pi $, where $n\in Z$.
So, let us calculate LHD and RHD of $g\left( x \right)=\csc x$ at point a from equation (i) and (ii)
Hence, LHD can be given as
LHD $=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,\dfrac{g\left( x \right)-g\left( a \right)}{x-a}$
Since, $g\left( x \right)=\csc x$, so , $g\left( a \right)=\csc a$
Hence, we get
LHD $=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,\dfrac{\csc x-\csc a}{x-a}$
Now, we can replace ${{a}^{-}}$ by $\left( a-h \right)$ where $h\to 0$, so, we get
LHD $=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\csc \left( a-h \right)-\csc a}{a-h-a}$
or LHD $=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\csc \left( a-h \right)-\csc a}{-h}$
We know that $\csc x=\dfrac{1}{\sin x}$. Hence, we get
LHD $=\underset{h\to 0}{\mathop{\lim }}\,\ -\dfrac{1}{h}\left[ \dfrac{1}{\sin \left( a-h \right)}-\dfrac{1}{\sin a} \right]$
LHD $=\underset{h\to 0}{\mathop{\lim }}\,\ -\dfrac{1}{h}\left[ \dfrac{\sin a-\sin \left( a-h \right)}{\sin \left( a-h \right)\sin a} \right]$
Now, we can use trigonometric identity as
$\sin C-\sin D=2\sin \left( \dfrac{C-D}{2} \right)\cos \left( \dfrac{C+D}{2} \right)$
Hence, above equation becomes
LHD $=\underset{h\to 0}{\mathop{\lim }}\,\ -\dfrac{1}{h}\left[ \dfrac{2\sin \left( \dfrac{a-a+h}{2} \right)\cos \left( \dfrac{a+h}{2} \right)}{\sin \left( a-h \right)\sin a} \right]$
LHD $=\underset{h\to 0}{\mathop{\lim }}\,\ -\dfrac{1}{h}\left[ \dfrac{2\sin \left( \dfrac{h}{2} \right)\cos \left( \dfrac{2a-h}{2} \right)}{\sin \left( a-h \right)\sin a} \right]$
or
LHD $=\underset{h\to 0}{\mathop{\lim }}\,\ -\dfrac{\sin \left( \dfrac{h}{2} \right)}{\left( \dfrac{h}{2} \right)}\left( \dfrac{\cos \left( \dfrac{2a-h}{2} \right)}{\sin \left( a-h \right)\sin a} \right)$
Now, we can use relation
$\underset{x\to 0}{\mathop{\lim }}\,\ \dfrac{\sin x}{x}=1$ with $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( \dfrac{h}{2} \right)}{\dfrac{h}{2}}$,
We get after applying limit $h\to 0$
LHD $=-1\dfrac{\cos a}{\sin a\cos a}$
We know that $\dfrac{\cos a}{\sin a}$ $=$$\cot a$and $\dfrac{1}{\sin a}=\csc a$, Hence, we get
LHD $=-\cot a\csc a$ ………………………………………………(iv)
Now we can calculate RHD from equation (ii), we get
RHD $=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,\dfrac{g\left( x \right)-g\left( a \right)}{x-a}$
As we have $g\left( x \right)=\csc x$, so $g\left( a \right)=\csc a$
Hence, we get
RHD $=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,\dfrac{\csc x-\csc a}{x-a}$
Now replace ${{a}^{+}}$ by $a+h$ where $h\to 0$
Hence, we get
RHD $=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{-\csc \left( a+h \right)-\csc a}{a+h-a}$
or RHD $=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{-\csc \left( a+h \right)-\csc a}{h}$
We know that $\csc x=\dfrac{1}{\sin x}$, Hence, we get
RHD $=\underset{h\to 0}{\mathop{\lim }}\,\ \dfrac{\left[ \dfrac{1}{\sin \left( a+h \right)}-\dfrac{1}{\sin a} \right]}{h}$
RHD $=\underset{h\to 0}{\mathop{\lim }}\,\ \dfrac{1}{h}\left[ \dfrac{\sin a-\sin \left( a+h \right)}{\sin \left( a+h \right)\sin a} \right]$
Now, we can use trigonometric identity as
$\sin C-\sin D=2\sin \left( \dfrac{C-D}{2} \right)\cos \left( \dfrac{C+D}{2} \right)$
Hence, RHD can be re-written as
RHD $=\underset{h\to 0}{\mathop{\lim }}\,\ \dfrac{1}{h}\left[ \dfrac{2\sin \left( \dfrac{a-a-h}{2} \right)\cos \left( \dfrac{a+a+h}{2} \right)}{\sin \left( a+h \right)\sin a} \right]$
RHD $=\underset{h\to 0}{\mathop{\lim }}\,\ \dfrac{1}{h}\left[ \dfrac{2\sin \left( \dfrac{-h}{2} \right)\cos \left( \dfrac{2a+h}{2} \right)}{\sin \left( a+h \right)\sin a} \right]$
or
RHD $=\underset{h\to 0}{\mathop{\lim }}\,\ -\left[ \left( \dfrac{\sin \left( \dfrac{h}{2} \right)}{\left( \dfrac{h}{2} \right)} \right)\dfrac{\cos \left( \dfrac{2a+h}{2} \right)}{\sin \left( a+h \right)\sin a} \right]$
where, we know $\sin \left( -x \right)=-\sin x$
Now, using the relation $\underset{x\to 0}{\mathop{\lim }}\,\ \dfrac{\sin x}{x}=1$, we get after putting limit $h\to 0$ to RHD;
RHD $=-\left[ \left( 1 \right)\dfrac{\cos a}{\sin a\sin a} \right]$
Hence, RHD $=-\cot a\csc a$ …………………………………………(v)
As, it is given that $g\left( x \right)=-\cot x\csc x$ , and hence$g'\left( a \right)=-\cot a\csc a$, Therefore, we get
LHD = RHD =$g'\left( a \right)$
Hence, the given function$g\left( x \right)=\csc x$ is differentiable at $x=a$ from equation (iii) where $a$$\ne $ $n\pi $.
Note: Don’t get confused with the term statement ‘$x$$\ne $ $n\pi $’ or ‘a is not multiple of $\pi $’. It is used because we cannot put $x=n\pi $ in $\csc x$. It will give positive, infinite or negative for $x\to n{{\pi }^{+}}$ or $x\to n{{\pi }^{-}}$. Hence $\csc x$ is not continuous at $x=n\pi $. That’s why we cannot put $x=n\pi $.
One can get confused with the identity $\sin C-\sin D$ , so be clear with the trigonometric identities with these kinds of questions.
One can use the L' Hospital rule for calculating LHD and RHD as LHD and RHD are of the form $\dfrac{0}{0}$. So, we need to use identities; we can use L’ Hospital as well.