How do you find f'(x) using the definition of a derivative for #(sqrt2x) - x^3#?

Question

Elijah Abram · Accepted Answer

#f'(x)=1/sqrt(2x)-3x^2# #f'(x)=lim_(h->0) (f(x+h)-f(x))/h# #f'(x)=lim_(h->0) (sqrt(2(x+h))-(x+h)^3-sqrt(2x)+x^3)/h# #f'(x)=lim_(h->0) (sqrt(2(x+h))-sqrt(2x))/h+lim_(h->0) (-(x+h)^3+x^3)/h# #f'(x)=lim_(h->0) (sqrt(2(x+h))-sqrt(2x))/h(sqrt(2(x+h))+sqrt(2x))/(sqrt(2(x+h))+sqrt(2x))+lim_(h->0) (-x^3 -3x^2h-3xh^2 -h^3+x^3)/h# #f'(x)=lim_(h->0) (2(x+h)-2x)/(h(sqrt(2(x+h))+sqrt(2x)))+lim_(h->0) (-x^3 -3x^2h-3xh^2 -h^3+x^3)/h# #f'(x)=lim_(h->0) (2(x+h)-2x)/(h(sqrt(2(x+h))+sqrt(2x)))+lim_(h->0) (-3x^2h-3xh^2 -h^3)/h# #f'(x)=lim_(h->0) (2x+2h-2x)/(h(sqrt(2(x+h))+sqrt(2x)))+lim_(h->0) (-3x^2-3xh -h^2)# #f'(x)=lim_(h->0) (2h)/(h(sqrt(2(x+h))+sqrt(2x))) - 3x^2# #f'(x)=lim_(h->0) 2/(sqrt(2(x+h))+sqrt(2x)) - 3x^2# #f'(x)=2/(2sqrt(2x)) - 3x^2=1/sqrt(2x)-3x^2#

Christopher Agresta · Answer

undefined To find $ f'(x) $ using the definition of a derivative for $ \sqrt{2x} - x^3 $, we apply the definition of the derivative:

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

First, we find $ f(x) $:

$$ f(x) = \sqrt{2x} - x^3 $$

Now, let's compute $ f(x+h) $:

$$ f(x+h) = \sqrt{2(x+h)} - (x+h)^3 $$

Next, we substitute $ f(x) $ and $ f(x+h) $ into the definition of the derivative:

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

Finally, we simplify the expression and compute the limit as $ h $ approaches zero.