# Differentiation

• Feb 8th 2010, 11:17 PM
Awsom Guy
Differentiation
Find the slope of the curve y= (x)/(x^2+1) at the origin. Find the equation of the tangent and of the normal at the origin.

Is this how I find y:
y= (0)/(0^2+1)
=0
where I have place x=0

I cannot attempt this as I have no idea where to start this question. I think I do not understand this question. Thanks.
• Feb 8th 2010, 11:21 PM
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Quote:

Originally Posted by Awsom Guy
Find the slope of the curve y= (x)/(x^2+1) at the origin. Find the equation of the tangent and of the normal at the origin.

Is this how I find y:
y= (0)/(0^2+1)
=0
where I have place x=0

I cannot attempt this as I have no idea where to start this question. I think I do not understand this question. Thanks.

You should know how to find the equation of a tangent.

Find the derivative, evaluate it at x = 0. Substitute into y = mx + c to find the y intercept.

For the normal, remember that $m_1m_2 = -1$ for perpendicular lines.
• Feb 8th 2010, 11:27 PM
Awsom Guy
If this is the derivation: (x^2-1)/(x^2+1)^2
then if I nake x=0 and do this:
(0-1)/(0^2+1)(0^2+1)
=-1/1
=-1
Does this find the y co-ordinate or the gradient.
• Feb 8th 2010, 11:39 PM
Prove It
Quote:

Originally Posted by Awsom Guy
If this is the derivation: (x^2-1)/(x^2+1)^2
then if I nake x=0 and do this:
(0-1)/(0^2+1)(0^2+1)
=-1/1
=-1
Does this find the y co-ordinate or the gradient.

Close, but the derivative is actually $\frac{1 - x^2}{(x^2 + 1)^2}$ (found using the Quotient Rule).

After you substitute $x = 0$, you have the GRADIENT of the tangent.
• Feb 8th 2010, 11:46 PM
Awsom Guy
Now how do I place that onto the y=mx+b formula, the gradient is 1. So it is y=1x+b. How do i find y, do I make y=0.
Thanks
• Feb 8th 2010, 11:52 PM
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Quote:

Originally Posted by Awsom Guy
Now how do I place that onto the y=mx+b formula, the gradient is 1. So it is y=1x+b. How do i find y, do I make y=0.
Thanks

You have already found that the point $(x, y) = (0, 0)$ lies on the curve. If the gradient is $1$, then $m = 1$.

Put these three values into the equation $y = mx + c$ and find $c$.
• Feb 8th 2010, 11:55 PM
Awsom Guy
oh ok thanks
• Feb 9th 2010, 12:13 AM
Awsom Guy
so that maks it y=1x
• Feb 9th 2010, 12:17 AM
Awsom Guy
just checking something, to calculate the normal do I inverse it and make it negative in this case 1/-x but just so you know it isn't 1/-x it is just -x.
• Feb 9th 2010, 12:18 AM
Awsom Guy
and how did you find x=0 and y=0, thanks
• Feb 9th 2010, 01:03 AM
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Quote:

Originally Posted by Awsom Guy
and how did you find x=0 and y=0, thanks