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

Question

Samantha Adkison · Accepted Answer

# f'(x) = 3x^2 +2x#

The definition of the derivative of #y=f(x)# is

# f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h #

So with # f(x) = x^3+x^2 # then;

# f(x+h) = (x+h)^3+(x+h)^2 # # \ \ \ \ \ \ \ = (x^3+3x^2h+3xh^2+h^3) +(x^2+2xh+h^2) #

And so:

# f(x+h)-f(x) = x^3+3x^2h+3xh^2+h^3 +x^2+2xh+h^2 -x^3-x^2 # # \ \ \ \ \ \ \ = 3x^2h+3xh^2+h^3 +2xh+h^2 #

And so the derivative of #f(x)# is given by:

# f'(x) = lim_(h rarr 0) ( f(x+h)-f(x) ) / h # # \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( 3x^2h+3xh^2+h^3 +2xh+h^2 ) / h # # \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( 3x^2+3xh+h^2 +2x+h )# # \ \ \ \ \ \ \ \ = 3x^2 +2x#

Ezekiel Alexander · Answer

undefined To find the derivative of $ f(x) = x^3 + x^2 $, apply the definition of the derivative:

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

Substitute the given function:

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

Expand and simplify the expression:

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

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

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

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

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