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

Answer 1

Short answer: use #a^3-b^3 = (a-b)(a^2+ab+b^2)# to "rationalize" the numerator.

We know that #a^2-b^2 = (a-b)(a+b)# and we use it to "rationalize" a difference of square roots. As in: #2/(3-sqrt5) = (2(3+sqrt5))/(9-5)#
For the crucial step in this problem, we use the fact that #a^3-b^3 = (a-b)(a^2+ab+b^2)#. This allows us to "rationalize" a difference of cube roots.
#a-b = (root(3)a - root(3)b)(root(3)a ^2+root(3)a root(3)b+root(3)b^2)#

In this case we have cube roots of squares, but the idea is the same.

From the identity above, we get:

#(a^(2/3)-b^(2/3)) (a^(4/3)+(ab)^(2/3)+b^(4/3)) = a^2-b^2#
#((x+h-6)^(2/3)-(x-6)^(2/3)) ((x+h-6)^(4/3)+((x+h-6)(x-6))^(2/3)+(x-6)^(4/3))#
# = (x+h-6)^2-(x-6)^2#
# = ((x-6)+h)^2 - (x-6)^2#
# = (x-6)^2+2(x-6)h +h^2 - (x-6)^2#
# = 2h(x-6)#

Using the above algebra, we can find:

#lim_(hrarr0)((x+h-6)^(2/3)-(x-6)^(2/3))/h#
#= lim_(hrarr0)(2h(x-6))/(h((x+h-6)^(4/3)+((x+h-6)(x-6))^(2/3)+(x-6)^(4/3)))#
# = (2(x-6))/(3(x-6)^(4/3)) = 2/(3(x-6)^(1/3))#
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Answer 2

To find ( f'(x) ) using the definition of a derivative for ( f(x) = (x - 6)^{2/3} ), you can follow these steps:

  1. Start with the definition of the derivative: [ f'(x) = \lim_{{h \to 0}} \frac{{f(x + h) - f(x)}}{h} ]

  2. Substitute the function ( f(x) ) into the definition: [ f'(x) = \lim_{{h \to 0}} \frac{{((x + h) - 6)^{2/3} - (x - 6)^{2/3}}{h} ]

  3. Expand the terms: [ f'(x) = \lim_{{h \to 0}} \frac{{((x + h) - 6)^{2/3} - (x - 6)^{2/3}}{h} ]

  4. Apply the binomial expansion formula: [ (a + b)^{2/3} = a^{2/3} + \frac{2}{3}a^{-1/3}b + \frac{1}{3}a^{-2/3}b^2 ]

  5. Substitute the expressions into the expanded form: [ f'(x) = \lim_{{h \to 0}} \frac{{((x - 6) + h)^{2/3} - (x - 6)^{2/3}}{h} ]

  6. Apply the binomial expansion formula to each term: [ ((x - 6) + h)^{2/3} = (x - 6)^{2/3} + \frac{2}{3}(x - 6)^{-1/3}h + \frac{1}{3}(x - 6)^{-2/3}h^2 ]

  7. Substitute the expressions into the derivative formula: [ f'(x) = \lim_{{h \to 0}} \frac{{(x - 6)^{2/3} + \frac{2}{3}(x - 6)^{-1/3}h + \frac{1}{3}(x - 6)^{-2/3}h^2 - (x - 6)^{2/3}}{h} ]

  8. Simplify and cancel terms: [ f'(x) = \lim_{{h \to 0}} \frac{{\frac{2}{3}(x - 6)^{-1/3}h + \frac{1}{3}(x - 6)^{-2/3}h^2}}{h} ] [ f'(x) = \lim_{{h \to 0}} \frac{{\frac{2}{3}h + \frac{1}{3}(x - 6)^{-1/3}h^2}}{h} ]

  9. Cancel the ( h ) terms: [ f'(x) = \lim_{{h \to 0}} \frac{{\frac{2}{3} + \frac{1}{3}(x - 6)^{-1/3}h}}{1} ]

  10. Evaluate the limit as ( h ) approaches 0: [ f'(x) = \frac{2}{3} ]

So, the derivative ( f'(x) ) of ( f(x) = (x - 6)^{2/3} ) is ( \frac{2}{3} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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