How do you find the derivative using limits of #h(x)=3+2/3x#?

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Answer 1 · 2017-09-21T09:54:32

see below

to avoid confusion with the #h# in the definition

we will replace #h(x)# with #f(x)#

#f'(x)=lim_(hrarr0)((f(x+h)-f(x))/h)#

we will replace #h(x)# with #f(x)#

#f(x)=3+2/3x#

#f'(x)=lim_(hrarr0)((3+2/3(x+h)-(3+2/3x))/h)#

#f'(x)=lim_(hrarr0)((cancel(3)+cancel(2/3x)+2/3hcancel(-3)cancel(-2/3x))/h)#

#f'(x)=lim_(hrarr0)((2/3cancel(h))/cancel(h))#

#f'(x)=2/3#