1. ## slope of tangent

$\displaystyle y=x-x^5$

find the slope of the tangent line to the curve at point (1,0)

slope =

$\displaystyle \frac{f(x)-f(a)}{x-a}$

I'm a little confused on what 'a' equals, but going off my notes im gonna say its 1, being x value in the point

$\displaystyle \frac{(x-x^5)-(1-1^5)}{x-1}$

=

$\displaystyle \frac{x-x^5}{x-1}$

and thats where I get stuck

2. Originally Posted by silencecloak
$\displaystyle y=x-x^5$

find the slope of the tangent line to the curve at point (1,0)

slope =

$\displaystyle \frac{f(x)-f(a)}{x-a}$

I'm a little confused on what 'a' equals, but going off my notes im gonna say its 1, being x value in the point

$\displaystyle \frac{(x-x^5)-(1-1^5)}{x-1}$

=

$\displaystyle \frac{x-x^5}{x-1}$

and thats where I get stuck
factor out the x, you get

$\displaystyle \lim_{x \to 1} \frac {x(1 - x^4)}{x - 1}$

now use the difference of two squares for the numerator

3. Originally Posted by Jhevon
factor out the x, you get

$\displaystyle \lim_{x \to 1} \frac {x(1 - x^4)}{x - 1}$

now use the difference of two squares for the numerator
$\displaystyle \lim_{x \to 1} \frac { x(1-x)(1+x)(1+x^2)}{x - 1}$

=

$\displaystyle \lim_{x \to 1} \frac{x-(x-1)(1+x)(1+x^2)}{x-1}$

=

$\displaystyle \lim_{x \to 1} \ x-(1+x)(1+x^2)$

=

$\displaystyle \lim_{x \to 1} 1-(1+1)(1+1)$

=

$\displaystyle \lim_{x \to 1} 1-(1+1)(1+1)$

= -3

pretty sure i goofed somewhere or im doing it wrong

4. Originally Posted by silencecloak
$\displaystyle \lim_{x \to 1} \frac { x(1-x)(1+x)(1+x^2)}{x - 1}$

=

$\displaystyle \lim_{x \to 1} \frac{x{\color{red}-}(x-1)(1+x)(1+x^2)}{x-1}$
well, yeah. obviously if you put stuff there that's not supposed to be there that will happen exactly where did that minus sign come from? do you realize that you are changing the value of the function?

5. Originally Posted by Jhevon
well, yeah. obviously if you put stuff there that's not supposed to be there that will happen exactly where did that minus sign come from? do you realize that you are changing the value of the function?
Ya, I was trying to pull a negative out so I could cancel out the denominator

EDIT:

durrr

$\displaystyle \frac{-x+1}{x-1} = -1$

so

$\displaystyle -x(1+x)(1+x^2) = -4$

6. Originally Posted by silencecloak
Ya, I was trying to pull a negative out so I could cancel out the denominator

EDIT:

durrr

$\displaystyle \frac{-x+1}{x-1} = -1$

so

$\displaystyle -x(1+x)(1+x^2) = -4$
yes

of course you have to write the limit until you actually take the limit