`h(s) = 3 + (2/3)s` Find the derivative of the function by the limit process.

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By limit process, the derivative of a function f(x) is :-

f'(x) = lim h --> 0 [{f(x+h) - f(x)}/h]

Now, the given function is :-

h(s) = 3 + (3/2)s

Thus, h'(s) = lim t --> 0 [{h(s+t) - h(t)}/t]

or, h'(s) = lim t ---> 0 [{{3 + (2/3)*(s+t)}...

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By limit process, the derivative of a function f(x) is :-

f'(x) = lim h --> 0 [{f(x+h) - f(x)}/h]

Now, the given function is :-

h(s) = 3 + (3/2)s

Thus, h'(s) = lim t --> 0 [{h(s+t) - h(t)}/t]

or, h'(s) = lim t ---> 0 [{{3 + (2/3)*(s+t)} - {3 + (2/3)s}}/t] 

or, h'(s) = lim t ---> 0 [{(2/3)t}/t] = 2/3

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