To find weather the function f(x) is derivable at a point x = c we have to check that

f ‘(c ^- ) = f ‘(c ⁺ ) = finite quantity, this condition must be fulfilled in order the function to be derivable.

As discussed above the top of this document where the description of the topic is given.

The given function is,

and we have to find weather it is derivable at x = c or not, so by using the formula,

f ‘(c) = , we get,

L.H.D. is given by,

f ‘(c ^- ) =

f ‘(c ^- ) =

f ‘(c ^- ) =

put x = c - h,

f ‘(c ^- ) =

f ‘(c ^- ) = lim_h→0 cos()

as the angle of cos is going to infinity, so it’s limit will oscillate between - 1 and 1, i.e. within the range of cos .

R.H.D. is given by,

f ‘(c ⁺ ) =

f ‘(c ⁺ ) =

f ‘(c ⁺ ) =

f ‘(c ⁺ ) = lim_h→0 cos()

as the angle of cos e is going to infinity, so it’s limit will oscillate between - 1 and 1 i.e. within the range of cos e.

As the results are also not equal to finite quantity, so hence the function is non-derivable.

Now we have to check the function for the continuity,

For a function to be continuous it should follow one condition fully, which is,

L.H.L = R.H.L. = f(x = c), at any point x = c.

Which can be further elaborated as,

= f(x = c)

Now we have to find L.H.L.,

f (c ^- ) =

put x = c - h,

f (c ^- ) = lim_h→0 (c - h - c)cos()

f (c ^- ) = lim_h→0 ( - h)cos()

As the cose function’s angle is approaching to infinity so it’s value will be any between - 1 to 1 and it is multiplied by h which is approaching to zero, so the whole limit will approach to zero.

f (c ^- ) = 0

Now we have to find R.H.L.,

f (c ⁺ ) =

put x = c + h,

f (c ⁺ ) =

f (c ⁺ ) = lim_h→0 (h)cos()

As the cose function’s angle is approaching to infinity so it’s value will be any between - 1 to 1 and it is multiplied by h which is approaching to zero, so the whole limit will approach to zero.

f (c ⁺ ) = 0

Hence as the L.H.L = R.H.L. = f(x = c) = 0

So our function is continuous but is non derivable.