1. ## [SOLVED] derivative

y=x^2/x+1 find all x values at which the tangent line is parallel to the line y=x

after you find the quotient of the problem how do you determine if its parallel to y=x

im at x^2+2x-1/x^2+2x+1

2. Originally Posted by vinson24
y=x^2/x+1 find all x values at which the tangent line is parallel to the line y=x

after you find the quotient of the problem how do you determine if its parallel to y=x

im at x^2+2x-1/x^2+2x+1
I get that $\frac{\,dy}{\,dx}=\frac{2x(x+1)-x^2}{(x+1)^2}=\frac{x^2+2x}{(x+1)^2}$.

Now find a x that causes $\frac{\,dy}{\,dx}=1$

Thus, $\frac{x^2+2x}{x^2+2x+1}=1$

Can this ever be true?

--Chris

3. sorry the numerator is x^2+1 i forgot to put the one in

4. Originally Posted by vinson24
sorry the numerator is x^2+1 i forgot to put the one in
It changes things a little:

$\frac{\,dy}{\,dx}=\frac{2x(x+1)-(x^2+1)}{(x+1)^2}=\frac{x^2+2x-1}{(x+1)^2}$

Find the value of x that causes $\frac{\,dy}{\,dx}=1$

This implies that $\frac{x^2+2x-1}{(x+1)^2}=1\implies x^2+2x-1=x^2+2x+1$

Can this ever be true?

--Chris

5. does that equal zero

6. Originally Posted by Chris L T521
It changes things a little:

$\frac{\,dy}{\,dx}=\frac{2x(x+1)-(x^2+1)}{(x+1)^2}=\frac{x^2+2x-1}{(x+1)^2}$

Find the value of x that causes $\frac{\,dy}{\,dx}=1$

This implies that $\frac{x^2+2x-1}{(x+1)^2}=1\implies x^2+2x-1=x^2+2x+1$

Can this ever be true?

--Chris
Originally Posted by vinson24
does that equal zero
Um...no.

$x^2+2x-1=x^2+2x+1\implies -1=1$

This is a false statement. Thus, there is no tangent line to this graph parallel to the line $y=x$.

--Chris

7. ooh because in the back of the book the answer it gave was 0

along those lines if you wanted dy/dx to pass through the orgin what does it equal to

8. i think by zero it means there isnt any thanks alot