Using the limit definition, how do you find the derivative of # f ( x) = x^4#?

Question

Christopher Agresta · Accepted Answer

#f' (x)=4x^3#

Given #f (x)=x^4#

Let #y=x^4#

replace #y# with #y+Delta y# and #x# with #x+Delta x#

#y=x^4#

#y+Delta y=(x+Delta x)^4#

#y+Delta y=x^4+4*x^3(Delta x)+6*x^2(Delta x)^2+4x(Delta x)^3+(Delta x)^4#

Subtract #y# from both sides

#y+Delta ycolor(red)(-y)=x^4+4*x^3(Delta x)+6*x^2(Delta x)^2+4x(Delta x)^3+(Delta x)^4color(red)(-x^4)#

#Delta y=4*x^3(Delta x)+6*x^2(Delta x)^2+4x(Delta x)^3+(Delta x)^4#

Divide both sides by #Delta x#

#(Delta y)/(Delta x)=(4*x^3(Delta x)+6*x^2(Delta x)^2+4x(Delta x)^3+(Delta x)^4)/(Delta x)#

#(Delta y)/(Delta x)=4*x^3+6*x^2(Delta x)+4x(Delta x)^2+(Delta x)^3#

Take now the #color(blue)("LIMIT")# of # (Delta y)/(Delta x)# as #Delta x rarr 0 #

#color(blue)(dy/dx=lim_(Deltax rarr 0)(Delta y)/(Delta x)=color(blue)(lim_(Delta xrarr 0)(4*x^3+6*x^2(Delta x)+4x(Delta x)^2+(Delta x)^3)=4x^3)#

God bless....I hope the explanation is useful.

Charles Arocha · Answer

Here is an alternative using one form of the limit definition.

I am using the limit definition in the form #f'(a) = lim_(xrarra)(f(x)-f(a))/(x-a)# For #f(x)=x^4#, we get #f'(a) = lim_(xrarra)(x^4-a^4)/(x-a)# #" "# (This has form #0/0#, so #x-a# is a factor of the numerator) # = lim_(xrarra)((x-a)(x^3+x^2a+xa^2+a^3))/(x-a)# #" "# (form #0/0#) # = lim_(xrarra)(x^3+x^2a+xa^2+a^3)# # = (a)^3+(a)^2a+(a)a^2+a^3# # = 4a^3# Since #f'(a) = 4a^3#, we get #f'(x) = 4x^3#

Emilia Aguilar · Answer

undefined To find the derivative of the function $f(x) = x^4$ using the limit definition of the derivative, we use the formula:

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

Substitute $f(x) = x^4$ into this formula:

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

Expand $(x + h)^4$:

$$(x + h)^4 = x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4$$

Substitute this expansion into the formula:

$$f'(x) = \lim_{h \to 0} \frac{(x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4) - x^4}{h}$$

Cancel $x^4$ terms:

$$f'(x) = \lim_{h \to 0} \frac{4x^3h + 6x^2h^2 + 4xh^3 + h^4}{h}$$

Factor out $h$ from the numerator:

$$f'(x) = \lim_{h \to 0} \frac{h(4x^3 + 6x^2h + 4xh^2 + h^3)}{h}$$

Cancel $h$ from the numerator and denominator:

$$f'(x) = \lim_{h \to 0} (4x^3 + 6x^2h + 4xh^2 + h^3)$$

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

$$f'(x) = 4x^3$$

Therefore, the derivative of the function $f(x) = x^4$ is $f'(x) = 4x^3$.