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

Question

Scarlett Allison · Accepted Answer

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#

Adam Adkins · Answer

undefined To find the derivative of the function $ h(x) = 3 + \frac{2}{3}x $ using limits, we use the definition of the derivative:

$$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $$

1. Substitute the function $ h(x) $ into the definition:

$$ h'(x) = \lim_{h \to 0} \frac{h(x + h) - h(x)}{h} $$

2. Replace $ h(x) $ with its expression:

$$ h'(x) = \lim_{h \to 0} \frac{3 + \frac{2}{3}(x + h) - (3 + \frac{2}{3}x)}{h} $$

3. Expand and simplify the expression:

$$ h'(x) = \lim_{h \to 0} \frac{3 + \frac{2}{3}x + \frac{2}{3}h - 3 - \frac{2}{3}x}{h} $$

$$ h'(x) = \lim_{h \to 0} \frac{\frac{2}{3}h}{h} $$

$$ h'(x) = \lim_{h \to 0} \frac{2}{3} $$

4. Since $ h $ is approaching 0, the limit becomes a constant:

$$ h'(x) = \frac{2}{3} $$

Therefore, the derivative of the function $ h(x) = 3 + \frac{2}{3}x $ is $ h'(x) = \frac{2}{3} $.