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  2. Solve! - Limit , continuity and differentiability - JEE Main-9
# Engineering

Right hand derivative of f(x)= \sin|x| + |\cos x| at x=\pi /2 equals

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Answers (1)
H Himanshu

 As we have learned

Right Hand Derivatives -

Right hand derivative of  f(x) at  x=x_{\circ }  is given by  

f'(x_{\circ })=\lim_{h\rightarrow 0}\:\:\frac{f(x_{\circ }+h)-f(x_{\circ })}{(x_{\circ }+h)-(x_{\circ })}

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 RHD = \lim_{h\rightarrow 0^{+}}\frac{f(\pi /2)+h-f(\pi /2)}{(\pi /2+h)-\pi /2}= \lim_{h\rightarrow 0^{+}}\frac{|\cos h|+|-\sin h|-1}{ h}

= \lim_{h\rightarrow 0^{+}}\frac{\cos h+-\sin h-1}{ h}=\lim_{h\rightarrow 0^{+}}\left ( \frac{\sin h}{h}-\frac{1- \cos h}{h} \right )

=\lim_{ h\rightarrow 0^{+}}\frac{\sin h}{h} - \lim_{h\rightarrow 0}^{+}\frac{1- \cos h}{h^{2}}\times h= 1-1/2*0=1

 

 

 

 

 

 


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