1. ## Differentiation

I'd like help with all of the first exercise, and then the two last parts of the last two exercises, 'cause I've done the first parts.

2. Find the equation of the normal line to
$\displaystyle y = \frac{1}{(x^2 + 1)^2}$
at the point
$\displaystyle \left ( 1, \frac{1}{4} \right )$

First we want the value of the slope of the tangent line at this point. So:
$\displaystyle \frac{dy}{dx} = -\frac{4x}{(x^2 + 1)^3}$

At x = 1 this is
$\displaystyle \frac{dy}{dx}_{x = 1} = -\frac{1}{2}$

Now, the line normal to the curve at this point has a slope of $\displaystyle -\frac{1}{-\frac{1}{2}} = 2$

So it has the form
$\displaystyle y = 2x + b$

We know this line passes through the point $\displaystyle \left ( 1, \frac{1}{4} \right )$, so
$\displaystyle \frac{1}{4} = 2 \cdot 1 + b$

$\displaystyle b = -\frac{7}{4}$

Thus the normal line is $\displaystyle y = 2x - \frac{7}{4}$

-Dan

3. Come now! The second part of number 3 is not a Calculus problem.

For what values of x is
$\displaystyle \frac{dy}{dx} = \frac{x^2 + 4x - 7}{(x + 2)^2}$
0 or undefined.

For the first part solve
$\displaystyle \frac{x^2 + 4x - 7}{(x + 2)^2} = 0$

For the second part, consider what values of x make the denominator 0 (which is forbidden.)

-Dan

4. Thanks for running me through the first one, I'm pretty sure I got what you did.

I'm not totally sure how to solve for 0...

5. Originally Posted by Fnus
Thanks for running me through the first one, I'm pretty sure I got what you did.

I'm not totally sure how to solve for 0...
Originally Posted by topsquark
For the first part solve
$\displaystyle \frac{x^2 + 4x - 7}{(x + 2)^2} = 0$
$\displaystyle \frac{x^2 + 4x - 7}{(x + 2)^2} = 0$

Multiply both sides by $\displaystyle (x + 2)^2$:
$\displaystyle x^2 + 4x - 7 = 0$

Can you solve this?

-Dan

6. Yep, I can.
Thanks a ton

- Frederikke