Use the definition of the derivative to find f’ when f(x) =6x^2 +4. No points for any other methods

please help??

Question

Use the definition of the derivative to find f’ when f(x) =6x^2 +4. No points for any other methods 
 
please help??

Isaiah Allen · Accepted Answer

#f'(x)=12x# According to definition of derivative for #f(x)#, #f'(x)=(df)/(dx)=lim_(deltax->0)(f(x+deltax)-f(x))/(deltax)# Here we have #f(x)=6x^2+4#, therefore #f(x+deltax)=6(x+deltax)^2+4# and #f'(x)=(df)/(dx)=lim_(deltax->0)(6(x+deltax)^2+4-(6x^2+4))/(deltax)# = #lim_(deltax->0)(6x^2+12xdeltax+6(deltax)^2+4-6x^2-4)/(deltax)# = #lim_(deltax->0)(12xdeltax+6(deltax)^2)/(deltax)# = #lim_(deltax->0)12x+6deltax# = #12x#

Christopher Agresta · Answer

#f'(x) = 12x# For any function #f#, the derivative #f'#, is given by: #f'(x) = lim_(hrarr0) (f(x+h) - f(x))/h# For #f(x) = 6x^2 + 4#, this is: #f'(x) = lim_(hrarr0) (6(x+h)^2 + 4 - (6x^2 +4))/h# #=lim_(hrarr0) (6x^2 + 12xh +6h^2 +4 - 6x^2 -4)/h# #= lim_(hrarr0) (12xh + 6h^2)/h# #=lim_(hrarr0)12x + 6h# Now it is just a matter of letting #h=0# which leaves: #f'(x) = 12x#

Adam Adkins · Answer

undefined To find $ f'(x) $ when $ f(x) = 6x^2 + 4 $, use the definition of the derivative:

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

Substitute $ f(x) = 6x^2 + 4 $ into the formula and simplify:

$$ f'(x) = \lim_{h \to 0} \frac{(6(x + h)^2 + 4) - (6x^2 + 4)}{h} $$

$$ f'(x) = \lim_{h \to 0} \frac{6(x^2 + 2hx + h^2) + 4 - 6x^2 - 4}{h} $$

$$ f'(x) = \lim_{h \to 0} \frac{6x^2 + 12hx + 6h^2 + 4 - 6x^2 - 4}{h} $$

$$ f'(x) = \lim_{h \to 0} \frac{12hx + 6h^2}{h} $$

$$ f'(x) = \lim_{h \to 0} (12x + 6h) $$

$$ f'(x) = 12x $$

Therefore, $ f'(x) = 12x $.

Use the definition of the derivative to find f’ when f(x) =6x^2 +4. No points for any other methods please help??