How do you find f'(x) using the definition of a derivative #f(x) =(x-6)^(2/3)#?

Question

Christopher Allers · Accepted Answer

Please see the explanation section below.

Use the fact that #a^3-b^3 = (a-b)(a^2+ab+b^2)#, so #a - b = (root3a-root3b)(root3a^2+root3(ab)+root3b^2)# Therefore the conjugate of #a^(2/3)-b^(2/3)# is #a^(4/3)+a^(2/3)b^(2/3)+b^(4/3)#. And the product is #a^2-b^2#. #lim_(hrarr0)((x+6+h)^(1/3)-(x+6)^(1/3))/h# To save some space, let's do the algebra first, then find the limit. #((x+6+h)^(2/3)-(x+6)^(2/3))/h = ((x+6+h)^2-(x+6)^2)/(h((x+6+h)^(4/3)+(x+6+h)^(2/3)(x+6)^(2/3)+(x+6)^(4/3))# # = (((x+6)+h)^2-(x+6)^2)/(h((x+6+h)^(4/3)+(x+6+h)^(2/3)(x+6)^(2/3)+(x+6)^(4/3))# # = ((x+6)^2+2(x+6)(h)+h^2-(x+6)^2)/(h((x+6+h)^(4/3)+(x+6+h)^(2/3)(x+6)^(2/3)+(x+6)^(4/3)))# # = (cancel(h)(2(x+6)+h))/(cancel(h)((x+6+h)^(4/3)+(x+6+h)^(2/3)(x+6)^(2/3)+(x+6)^(4/3)))# So, we have #lim_(hrarr0)((x+6+h)^(1/3)-(x+6)^(1/3))/h= lim_(hrarr0)(2(x+6)+h)/((x+6+h)^(4/3)+(x+6+h)^(2/3)(x+6)^(2/3)+(x+6)^(4/3))# # = (2(x+6))/((x+6)^(4/3)+(x+6)^(2/3)(x+6)^(2/3)+(x+6)^(4/3))# # = (2(x+6))/(3(x+6)^(4/3)# # = 2/(3(x+6)^(1/3))#

Scarlett Allison · Answer

Use #a^3 - b^3 = (a - b)(a^2 + ab + b^2)# as a pattern to clear the #1/3# power from the numerator. Please see the explanation.

Use: #f'(x) = lim_(hto0)(f(x+h)-f(x))/h# Given: #f(x) = (x - 6)^(2/3)#, then #f(x+h) = (x + h-6)^(2/3)# This notation can get hard to follow so I am going to make the following substitution: Let #a = f(x + h)# Let #b = f(x)# #f'(x) = lim_(hto0)(a-b)/h# We want the numerator to become #a^3 - b^3#, because that will change the power of each of the terms from #2/3# to 2. Therefore, we multiply by #(a^2 + ab + b^2)/(a^2 + ab + b^2)# #f'(x) = lim_(hto0)(a-b)/h(a^2 + ab + b^2)/(a^2 + ab + b^2)# The numerator becomes what we want: #f'(x) = lim_(hto0)(a^3-b^3)/(h(a^2 + ab + b^2))" [1]"# Now we work with the numerator so that we can make a factor of h cancel out the h in the denominator. #a^3 = (f(x + h))^3# #a^3 = ((x + h - 6)^(2/3))^3# #a^3 = (x + h - 6)^2# #a^3 = x^2 +hx -6x + hx + h^2 - 6h -6x -6h - 36" [2]"# #b^3 = (f(x))^3# #b^3 = ((x - 6)^(2/3))^3# #b^3 = (x - 6)^2# #b^3 = x^2 - 12x - 36" [3]"# Subtract equation [3] from equation [2]: #a^3 - b^3 = x^2 +hx -6x + hx + h^2 - 6h -6x -6h - 36 - x^2 + 12x + 36# #a^3 - b^3 = cancel(x^2 +)hx cancel(-6x) + hx + h^2 - 6h cancel(-6x) -6h cancel(- 36)cancel( - x^2) cancel(+ 12x)cancel( + 36)# #a^3 - b^3 = 2hx + h^2 - 12h# #a^3 - b^3 = h(2x + h - 12)" [4]"# Substitute the right side of equation [4] into the numerator of equation [1]: #f'(x) = lim_(hto0)(h(2x + h - 12))/(h(a^2 + ab + b^2))" [5]"# The h in the numerator cancels the h in the denominator: #f'(x) = lim_(hto0)(cancel(h)(2x + h - 12))/(cancel(h)(a^2 + ab + b^2))# We can let #hto 0#. This makes "a" become "b" so the denominator becomes #3b^2# #f'(x) = (2x - 12)/(3b^2)# Substitute f(x) for b: #f'(x) = (2x - 12)/(3(f(x))^2)# Remove a factor of 2 from the numerator: #f'(x) = 2(x - 6)/(3(f(x))^2)# Remove a factor of 3 from the denominator: #f'(x) = 2/3(x - 6)/(f(x))^2# Substitute #(x - 6)^(2/3)# for #f(x)# #f'(x) = 2/3(x - 6)/(((x - 6)^(2/3))^2)# #f'(x) = 2/3(x - 6)/((x - 6)^(4/3))# #f'(x) = 2/3(x - 6)^(-1/3)# This is what we expected.

Charles Anctil · Answer

undefined To find $ f'(x) $ using the definition of a derivative for $ f(x) = (x-6)^{2/3} $, you need to apply the limit definition of the derivative:

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

$$ = \lim_{h \to 0} \frac{((x+h)-6)^{2/3} - (x-6)^{2/3}}{h} $$