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

2 Answers
Jan 11, 2017

Please see the explanation section below.

Explanation:

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))#

Jan 11, 2017

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.

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.