How do you find the derivative using limits of $h (x) = 3 + \frac{2}{3} x$ $h(x)=3+2/3x$ ?

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Answer 1 · 2017-09-21T09:54:32

see below

to avoid confusion with the $h$ $h$ in the definition

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

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

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

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

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

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

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

$f'(x)=2/3$

How do you find the derivative using limits of h(x)=3+2/3xh(x)=3+23xh(x)=3+2/3x?