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Question
AnswersIf $a\ne \left( 2n+1 \right)\dfrac{\pi }{2}$, $n\in z$, then show that the function $h\left( x \right)=\sec x$ is differentiable at $a$ and $h'\left( a \right)=\sec a\tan a$. In general, $h'\left( x \right)=\sec x\tan x$ for all $x\ne \left( 2n+1 \right)\dfrac{\pi }{2}$,$n\in Z$.
Answer
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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$=\underset{x\to {{c}^{-}}}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( c \right)}{x-c}$ 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 its 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 any function $f\left( x \right)$ at point ‘c’ can be given as
LHD$=\underset{x\to {{c}^{-}}}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( c \right)}{x-c}$ …………………………………………………..(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, function $h\left( x \right)$ is given as $\sec x$ where $h'\left( x \right)=\sec x\tan x$ for all $x\ne \left( 2n+1 \right)\dfrac{\pi }{2},n\in z$.
And we need to determine whether the given function is differentiable at $x=a$ if $a\ne \left( 2n+1 \right)\dfrac{\pi }{2},n\in z$ and $h'\left( a \right)=\sec a\cdot \tan a$$h'\left( a \right)=\sec a\cdot \tan a$.
So, from equation (i) LHD can be given as
LHD$=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,\dfrac{h\left( x \right)-h\left( a \right)}{x-a}$
As, $h\left( x \right)=\sec x$ and hence $h\left( a \right)=\sec a$. So, LHD can be written as
LHD$=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,\dfrac{\sec x-\sec a}{x-a}$
Now, we can replace ‘a’ by ‘a-h’ where $h\to 0$. Hence, above equations can be written in ‘h’ as
LHD$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sec \left( a-h \right)-\sec a}{a-h-a}$
or
$\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sec \left( a-h \right)-\sec a}{-h}$
Now, we know that $\sec x=\dfrac{1}{\cos x}$ ; Hence, we get,
LHD$=\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{1}{\cos \left( a-h \right)}-\dfrac{1}{\cos a} \right)\left( \dfrac{-1}{h} \right)$
LHD$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{-1}{h}\left[ \dfrac{\cos a-\cos \left( a-h \right)}{\cos \left( a-h \right)\cos a} \right]$
Now, we can apply trigonometry identity of $\cos C-\cos D$ which is given as :-
$\cos C-\cos D=-2\sin \left( \dfrac{C-D}{2} \right)\sin \left( \dfrac{C+D}{2} \right)$
Hence, LHD can be written as
LHD$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{-1}{h}\left[ \dfrac{-2\sin \dfrac{a-a+h}{2}\sin \dfrac{a+a-h}{2}}{\cos \left( a-h \right)\cos a} \right]$
LHD$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{-1}{h}\left[ \dfrac{-2\sin \dfrac{h}{2}\sin \dfrac{2a-h}{2}}{\cos \left( a-h \right)\cos a} \right]$
or
LHD$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\left( \sin \dfrac{h}{2} \right)}{\left( \dfrac{h}{2} \right)}\times \sin \dfrac{\left( \dfrac{2a-h}{2} \right)}{\cos \left( a-h \right)\cos a}$
Now, we can relation of $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1$ , hence we get
LHD$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\left( 1 \right)\sin \left( \dfrac{2a-h}{2} \right)}{\cos \left( a-h \right)\cos a}$
On applying limits, we get
LHD$=\dfrac{\sin a}{\cos a\cos a}$
Now, we know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\sec \theta =\dfrac{1}{\cos \theta }$. Hence, LHD can be given as
LHD$=\sec a\tan a$ ……………………………………………………………….. (iv)
Now, we can calculate RHD by equation (ii), we get
RHD$=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,\dfrac{h\left( x \right)-h\left( a \right)}{x-a}$
Now, we have $h\left( x \right)=\sec x$, hence, we have $h\left( a \right)=\sec a$. So, RHD can be written as
RHD$=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,\dfrac{\sec x-\sec a}{x-a}$
Now, replace ${{a}^{+}}$ by $\left( a+h \right)$ where $h\to 0$.
Hence, we get
RHD$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sec \left( a+h \right)-\sec a}{a+h-a}$
RHD$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sec \left( a+h \right)-\sec a}{h}$
Now, use $\sec x=\dfrac{1}{\cos x}$, we get
RHD$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{1}{\cos \left( a+h \right)}-\dfrac{1}{\cos a}}{h}$
RHD$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{h}\left[ \dfrac{\cos a-\cos \left( a+h \right)}{\cos \left( a+h \right)\cos a} \right]$
Now, use trigonometric identity of $\cos C-\cos D$ which is given as
$\cos C-\cos D=-2\sin \dfrac{C-D}{2}\sin \dfrac{C+D}{2}$
Hence, RHD can be given as
RHD$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{h}\left[ \dfrac{-2\sin \left( \dfrac{a-a-h}{2} \right)\sin \left( \dfrac{a+a+h}{2} \right)}{\cos \left( a+h \right)\cos a} \right]$
or
RHD$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{h}\left[ \dfrac{-2\sin \left( \dfrac{-h}{2} \right)\sin \left( \dfrac{2a+h}{2} \right)}{\cos \left( a+h \right)\cos a} \right]$
We know $\sin \left( -x \right)=-\sin x$, hence above relation can be given as
RHD$\underset{h\to 0}{\mathop{=\lim }}\,\dfrac{2}{h}\sin \left( \dfrac{h}{2} \right)\dfrac{\sin \left( \dfrac{2a+h}{2} \right)}{\cos \left( a+h \right)\cos a}$
or
$\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( \dfrac{h}{2} \right)}{\left( \dfrac{h}{2} \right)}\dfrac{\sin \left( \dfrac{2a+h}{2} \right)}{\cos \left( a+h \right)\cos a}$
Now, we can use the relation $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1$ to simplify the above equation. Hence, we get
RHD$=\underset{h\to 0}{\mathop{\lim }}\,\left( 1 \right)\dfrac{\sin \left( \dfrac{2a+h}{2} \right)}{\cos \left( a+h \right)\cos a}$
On putting limits to the above equation, we get
RHD$=\dfrac{\sin a}{\cos a\cos a}=\sec a\tan a$
Hence,
RHD$=\sec a\tan a$ ………………………………………………………………… (v)
Now, we can observe that LHD, RHD at point ‘a’ and value of derivative of $h\left( x \right)$ at point a, all are equal to $\sec a\tan a$. Hence, from equation (iii), we get to know that $h\left( x \right)=\sec x$ is differentiable at $x=a$ where $a\ne \left( 2n+1 \right)\dfrac{\pi }{2}$.
NOTE: Don’t confuse with the statement $x\ne \left( 2n+1 \right)\dfrac{\pi }{2}$ or a is not an odd multiple of$\dfrac{\pi }{2}$ . These are given because we can not put $x=\left( 2n+1 \right)\dfrac{\pi }{2}$ to function $\sec x$ as it will give positive infinite or negative for $\left( 2n+1 \right){{\dfrac{\pi }{2}}^{-}}$ or $\left( 2n+1 \right){{\dfrac{\pi }{2}}^{+}}$.
That’s why these statements are used in question.
We can use the L'Hospital Rule while calculating LHD and RHD values as both are of the form $\dfrac{0}{0}$. So, we don’t need to use any trigonometric identity for solving LHD and RHD if we use L’Hospital Rule.
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{x\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 its 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 any function $f\left( x \right)$ at point ‘c’ can be given as
LHD$=\underset{x\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, function $h\left( x \right)$ is given as $\sec x$ where $h'\left( x \right)=\sec x\tan x$ for all $x\ne \left( 2n+1 \right)\dfrac{\pi }{2},n\in z$.
And we need to determine whether the given function is differentiable at $x=a$ if $a\ne \left( 2n+1 \right)\dfrac{\pi }{2},n\in z$ and $h'\left( a \right)=\sec a\cdot \tan a$$h'\left( a \right)=\sec a\cdot \tan a$.
So, from equation (i) LHD can be given as
LHD$=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,\dfrac{h\left( x \right)-h\left( a \right)}{x-a}$
As, $h\left( x \right)=\sec x$ and hence $h\left( a \right)=\sec a$. So, LHD can be written as
LHD$=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,\dfrac{\sec x-\sec a}{x-a}$
Now, we can replace ‘a’ by ‘a-h’ where $h\to 0$. Hence, above equations can be written in ‘h’ as
LHD$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sec \left( a-h \right)-\sec a}{a-h-a}$
or
$\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sec \left( a-h \right)-\sec a}{-h}$
Now, we know that $\sec x=\dfrac{1}{\cos x}$ ; Hence, we get,
LHD$=\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{1}{\cos \left( a-h \right)}-\dfrac{1}{\cos a} \right)\left( \dfrac{-1}{h} \right)$
LHD$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{-1}{h}\left[ \dfrac{\cos a-\cos \left( a-h \right)}{\cos \left( a-h \right)\cos a} \right]$
Now, we can apply trigonometry identity of $\cos C-\cos D$ which is given as :-
$\cos C-\cos D=-2\sin \left( \dfrac{C-D}{2} \right)\sin \left( \dfrac{C+D}{2} \right)$
Hence, LHD can be written as
LHD$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{-1}{h}\left[ \dfrac{-2\sin \dfrac{a-a+h}{2}\sin \dfrac{a+a-h}{2}}{\cos \left( a-h \right)\cos a} \right]$
LHD$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{-1}{h}\left[ \dfrac{-2\sin \dfrac{h}{2}\sin \dfrac{2a-h}{2}}{\cos \left( a-h \right)\cos a} \right]$
or
LHD$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\left( \sin \dfrac{h}{2} \right)}{\left( \dfrac{h}{2} \right)}\times \sin \dfrac{\left( \dfrac{2a-h}{2} \right)}{\cos \left( a-h \right)\cos a}$
Now, we can relation of $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1$ , hence we get
LHD$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\left( 1 \right)\sin \left( \dfrac{2a-h}{2} \right)}{\cos \left( a-h \right)\cos a}$
On applying limits, we get
LHD$=\dfrac{\sin a}{\cos a\cos a}$
Now, we know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\sec \theta =\dfrac{1}{\cos \theta }$. Hence, LHD can be given as
LHD$=\sec a\tan a$ ……………………………………………………………….. (iv)
Now, we can calculate RHD by equation (ii), we get
RHD$=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,\dfrac{h\left( x \right)-h\left( a \right)}{x-a}$
Now, we have $h\left( x \right)=\sec x$, hence, we have $h\left( a \right)=\sec a$. So, RHD can be written as
RHD$=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,\dfrac{\sec x-\sec a}{x-a}$
Now, replace ${{a}^{+}}$ by $\left( a+h \right)$ where $h\to 0$.
Hence, we get
RHD$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sec \left( a+h \right)-\sec a}{a+h-a}$
RHD$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sec \left( a+h \right)-\sec a}{h}$
Now, use $\sec x=\dfrac{1}{\cos x}$, we get
RHD$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{1}{\cos \left( a+h \right)}-\dfrac{1}{\cos a}}{h}$
RHD$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{h}\left[ \dfrac{\cos a-\cos \left( a+h \right)}{\cos \left( a+h \right)\cos a} \right]$
Now, use trigonometric identity of $\cos C-\cos D$ which is given as
$\cos C-\cos D=-2\sin \dfrac{C-D}{2}\sin \dfrac{C+D}{2}$
Hence, RHD can be given as
RHD$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{h}\left[ \dfrac{-2\sin \left( \dfrac{a-a-h}{2} \right)\sin \left( \dfrac{a+a+h}{2} \right)}{\cos \left( a+h \right)\cos a} \right]$
or
RHD$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{h}\left[ \dfrac{-2\sin \left( \dfrac{-h}{2} \right)\sin \left( \dfrac{2a+h}{2} \right)}{\cos \left( a+h \right)\cos a} \right]$
We know $\sin \left( -x \right)=-\sin x$, hence above relation can be given as
RHD$\underset{h\to 0}{\mathop{=\lim }}\,\dfrac{2}{h}\sin \left( \dfrac{h}{2} \right)\dfrac{\sin \left( \dfrac{2a+h}{2} \right)}{\cos \left( a+h \right)\cos a}$
or
$\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( \dfrac{h}{2} \right)}{\left( \dfrac{h}{2} \right)}\dfrac{\sin \left( \dfrac{2a+h}{2} \right)}{\cos \left( a+h \right)\cos a}$
Now, we can use the relation $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1$ to simplify the above equation. Hence, we get
RHD$=\underset{h\to 0}{\mathop{\lim }}\,\left( 1 \right)\dfrac{\sin \left( \dfrac{2a+h}{2} \right)}{\cos \left( a+h \right)\cos a}$
On putting limits to the above equation, we get
RHD$=\dfrac{\sin a}{\cos a\cos a}=\sec a\tan a$
Hence,
RHD$=\sec a\tan a$ ………………………………………………………………… (v)
Now, we can observe that LHD, RHD at point ‘a’ and value of derivative of $h\left( x \right)$ at point a, all are equal to $\sec a\tan a$. Hence, from equation (iii), we get to know that $h\left( x \right)=\sec x$ is differentiable at $x=a$ where $a\ne \left( 2n+1 \right)\dfrac{\pi }{2}$.
NOTE: Don’t confuse with the statement $x\ne \left( 2n+1 \right)\dfrac{\pi }{2}$ or a is not an odd multiple of$\dfrac{\pi }{2}$ . These are given because we can not put $x=\left( 2n+1 \right)\dfrac{\pi }{2}$ to function $\sec x$ as it will give positive infinite or negative for $\left( 2n+1 \right){{\dfrac{\pi }{2}}^{-}}$ or $\left( 2n+1 \right){{\dfrac{\pi }{2}}^{+}}$.
That’s why these statements are used in question.
We can use the L'Hospital Rule while calculating LHD and RHD values as both are of the form $\dfrac{0}{0}$. So, we don’t need to use any trigonometric identity for solving LHD and RHD if we use L’Hospital Rule.