How do you find the derivative of #3x^2-5x+2# using the limit definition?

Question

Paisley Johnson · Accepted Answer

# f'(x)=6x-5 # By definition of the derivative # f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h # So with # f(x) = 3x^2 - 5x + 2 # we have; # f'(x)=lim_(h rarr 0) ( {3(x+h)^2-5(x+h)+2 } - {3x^2 - 5x + 2 } ) / h # # :. f'(x)=lim_(h rarr 0) ( {3(x^2+2hx+h^2)-5(x+h)+2 } - {3x^2 - 5x + 2 } ) / h # # :. f'(x)=lim_(h rarr 0) ( {3x^2+6hx+3h^2-5x-5h+2 - 3x^2 + 5x - 2 } ) / h # # :. f'(x)=lim_(h rarr 0) ( {6hx+3h^2-5h } ) / h # # :. f'(x)=lim_(h rarr 0) ( 6x+3h-5 ) # # :. f'(x)=6x-5 # # :. f'(x)=6x-5 #

Cameron Alexander · Answer

undefined To find the derivative of $3x^2 - 5x + 2$ using the limit definition, you would use the formula:

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

where $f(x) = 3x^2 - 5x + 2$.

Substitute $f(x + h)$ and $f(x)$ into the formula:

$$
f(x + h) = 3(x + h)^2 - 5(x + h) + 2
$$
$$
f(x) = 3x^2 - 5x + 2
$$

Expand $f(x + h)$ and simplify:

$$
f(x + h) = 3(x^2 + 2hx + h^2) - 5x - 5h + 2
$$
$$
f(x + h) = 3x^2 + 6hx + 3h^2 - 5x - 5h + 2
$$

Substitute $f(x + h)$ and $f(x)$ back into the formula and simplify:

$$
f'(x) = \lim_{{h \to 0}} \frac{{(3x^2 + 6hx + 3h^2 - 5x - 5h + 2) - (3x^2 - 5x + 2)}}{h}
$$
$$
f'(x) = \lim_{{h \to 0}} \frac{{6hx + 3h^2 - 5h}}{h}
$$

Factor out $h$ from the numerator:

$$
f'(x) = \lim_{{h \to 0}} \frac{{h(6x + 3h - 5)}}{h}
$$

Cancel out the $h$ terms:

$$
f'(x) = \lim_{{h \to 0}} (6x + 3h - 5)
$$

Substitute $h = 0$:

$$
f'(x) = 6x - 5
$$

Therefore, the derivative of $3x^2 - 5x + 2$ with respect to $x$ using the limit definition is $6x - 5$.