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

2 Answers

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

Explanation:

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.

Apr 20, 2016

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

Explanation:

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#