Lf fx- 1- x-0- 1- sinx- 0-x-2 - then derivative of fx at x-0

The correct option is D does not existLHD at x=0= [ d / dx 1 ] #160;_ x=0 =0 RHD at x=0= [ d / dx 1+sin x #160;] #160;_ x=0 =cos0=1 hence ...

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Graphical Interpretation of Differentiability

lf fx=[ ...

Question

lf $f (x) = {\begin{matrix} 1, x < 0 1 + s i n x, 0 \leq x < / π_{2} \end{matrix}$ , then derivative of f(x) at $x = 0$

is equal to 1

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is equal to 0

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is equal to -1

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does not exist

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Solution

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