How do you find the derivative of #f(x) = x^2+3x+1# using the limit definition?

Question

Charles Anctil · Accepted Answer

Do some substituting and algebra to get #f'(x)=2x+3#.

The definition of the derivative is: #f'(x)=lim_(h->0)(f(x+h)-f(x))/h# Our function #f(x)# equals #x^2+3x+1#. Applying it to the definition, we have: #f'(x)=lim_(h->0)((x+h)^2+3(x+h)+1-(x^2+3x+1))/h# And doing some algebra to finish off: #f'(x)=lim_(h->0)(x^2+2xh+h^2+3x+3h+1-x^2-3x-1)/h# #f'(x)=lim_(h->0)(2xh+h^2+3h)/h# #f'(x)=lim_(h->0)(h(2x+h+3))/h# #f'(x)=lim_(h->0)2x+h+3# #f'(x)=2x+(0)+3# #f'(x)=2x+3#

Isaiah Allen · Answer

undefined To find the derivative of $ f(x) = x^2 + 3x + 1 $ using the limit definition, you would start by applying the definition of the derivative, which is:

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

Substitute $ f(x) = x^2 + 3x + 1 $ into the formula:

$$ f'(x) = \lim_{h \to 0} \frac{(x + h)^2 + 3(x + h) + 1 - (x^2 + 3x + 1)}{h} $$

Simplify the expression inside the limit:

$$ f'(x) = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 + 3x + 3h + 1 - x^2 - 3x - 1}{h} $$

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

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

Now, as $ h $ approaches 0, the limit becomes:

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

So, the derivative of $ f(x) = x^2 + 3x + 1 $ is $ f'(x) = 2x + 3 $.