Calculate the slope of the curve at x=1

[Kazane]

Homework Statement

f(x)=x^1/3
use the definition of f'(a) to calculate the slope of the curve at x=1
(Hint: by rationalizing the numerator. useful formula a³ -b³=(a-b)(a²+ab+b²)

The Attempt at a Solution

f'(x)=lim (x->1) f(x)-f(1)/x-1

=lim (x->1) x^1/3-1^1/3/ x-1

How can I do x^1/3-1^1/3 to a³ -b³=(a-b)(a²+ab+b²)?

I'm confused so please help!

 

[Ray Vickson]

Homework Statement

f(x)=x^1/3
use the definition of f'(a) to calculate the slope of the curve at x=1
(Hint: by rationalizing the numerator. useful formula a³ -b³=(a-b)(a²+ab+b²)

The Attempt at a Solution

f'(x)=lim (x->1) f(x)-f(1)/x-1

=lim (x->1) x^1/3-1^1/3/ x-1

How can I do x^1/3-1^1/3 to a³ -b³=(a-b)(a²+ab+b²)?

I'm confused so please help!

You should never write f(x) - f(1)/x-1, which means f(x) - [f(1)/x] - 1 when evaluated according to standard, universally accepted rules. You should use brackets and write [f(x) - f(1)]/(x-1).

RGV

 

[HallsofIvy]

Homework Statement

f(x)=x^1/3
use the definition of f'(a) to calculate the slope of the curve at x=1
(Hint: by rationalizing the numerator. useful formula a³ -b³=(a-b)(a²+ab+b²)

The Attempt at a Solution

f'(x)=lim (x->1) f(x)-f(1)/x-1

=lim (x->1) x^1/3-1^1/3/ x-1

How can I do x^1/3-1^1/3 to a³ -b³=(a-b)(a²+ab+b²)?

I'm confused so please help!

Let $a= x^{1/3}$, $b= y^{1/3}$