# Finding the equations of tangents

• Mar 21st 2011, 11:12 AM
Googl
Finding the equations of tangents
Finding the equations of the tangent to the curve at the given point.

$\displaystyle y = 4/x^2 at$ x = 1

I understand how to work out this but I haven't figured out how to treat fractions. For example I can workout $\displaystyle y = 5x^2 + 3$.

Please help as this for a test and I need to figure out how to work it out.

Thanks.
• Mar 21st 2011, 11:16 AM
e^(i*pi)

$\displaystyle \dfrac{4}{x^2} = 4x^{-2}$
• Mar 22nd 2011, 10:30 AM
Googl
Quote:

Originally Posted by e^(i*pi)

$\displaystyle \dfrac{4}{x^2} = 4x^{-2}$

Thanks.

But how can I work out:

4(2+h)^2

I have tried working it work but I couldn't. I am trying to find the y axis value to replace (x, ?)
• Mar 22nd 2011, 10:48 AM
skeeter
use the alternate definition of a derivative ...

$\displaystyle \displaystyle f'(a) = \lim_{x \to a} \dfrac{f(x) - f(a)}{x-a}$

... where $\displaystyle a = 1$
• Mar 22nd 2011, 10:55 AM
HallsofIvy
Quote:

Originally Posted by Googl
Thanks.

But how can I work out:

4(2+h)^2

I have tried working it work but I couldn't. I am trying to find the y axis value to replace (x, ?)

You did not say that you have to find the derivative from the "difference quotient" formula.

I assume you mean $\displaystyle 4(2+ h)^{-2}$. To do that, I recommend you go back to the fraction form: $\displaystyle \frac{4}{x^2}$ so that the difference quotient is $\displaystyle \frac{\frac{4}{(2+h)^2}- \frac{4}{2^2}}{h}$.

To simplify that, get "common denominators". $\displaystyle \frac{4}{(2+h)^2}- 1= \frac{4}{(2+h)^2}- \frac{(2+h)^2}{(2+h) ^2}$$\displaystyle = \frac{4- (4+ 4h+ h^2)}{(2+h)^2}= \frac{-4h- h^2}{(2+ h)^2}$

Dividing that by h, $\displaystyle \frac{-4h- h^2}{h(2+ h)^2}= \frac{h(-4- h)}{h(2+h)^2}= \frac{-4-h}{(2+ h)^2}$

Now, take the limit of that as h goes to 0.
• Mar 22nd 2011, 07:17 PM
Googl
Hello HallsofIvy,

Your working out is not what I am used to. How you have managed to transform the fractions... And I have to learn how to transform it the way you have. I understand how to find the equations of the tangent to the curve at the given point. This is what I have been revising at the moment, but when it came to the y equalling to a fraction curve that's where I got stuck. My reference doesn't actually show how to work out this. It shows for the standard ones without fractions but in the practice question it has some of the questions as fractions.

I understand I am supposed to use y1 - y = m(x - x1) where m is the gradient. So I am supposed to find the gradient first which I know how to, and then y1 and x1 and replace them in the equation. That's how I am working it out at the moment.

In your reply I don't understand why you had to subtract by 1. And why you concentrated on just the denominator towards the end, also I didn't quite understand how the final step could be useful.

I think the best way for me to figure out how to handle these types of questions is to get the full working out for the $\displaystyle y = 4/x^2$ at x = 1 (finding the equations of the tangent to the curve at the given point.)

Thanks.