Question

How do you find the derivative by definition for `y=x^(7/3)`?

Answer 1

The details depend on whether you use

`f'(a) = lim_(xrarra) (x^(7/3) - a^(7/3))/(x-a)`

or,

`f'(a) = lim_(hrarr0) ( (a+h)^(7/3) - a^(7/3))/(h)`.

Either way, we'll start with the same algebra:

`u^3 - v^3 = (u-v) (u^2 +uv + v^2)`.

So we can 'rationalize' differences involving cube roots:

`u - v = (u^(1/3) -v^(1/3) ) (u^(2/3) +u^(1/3) v^(1/3) + v^(2/3) )`.

This may be easier to see written:

`u - v = (root(3)u-root(3)v ) (root(3)u^2 +root(3)(u v) + root(3)v^2 )`.

This allows us to rewrite the fraction with a numerator that is a difference of 7th powers.

Then, depending on the form of the definition you are using:

Factor:
`x^7 - a^7 = (x-a)(x^6+x^5a + x^4a^2 + x^3a^3 + x^2a^4 + xa^5 + a^6`.

Or, expand (using the binomial theorem)

`(a+h)^7 - a^7 = [a^7 + 7a^6h + ((7),(2))a^5h^2 + ((7),(3))a^4h^3 + * * * + 7ah^6 + h^7] - a^7`.

Finally, reduce the fraction and evaluate the limit.