# Thread: integration question

1. ## integration question

$\displaystyle y=\frac{4}{x^2}$

why is this

$\displaystyle y=\frac{4x}{x^3}$

and not

$\displaystyle y=\frac{4x}{0.3333x^3}$

sorry i dont know hwo to do a fraction in a fraction so 0.33333 = one third

2. Originally Posted by coyoteflare
$\displaystyle y=\frac{4}{x^2}$

why is this

$\displaystyle y=\frac{4x}{x^3}$

and not

$\displaystyle y=\frac{4x}{0.3333x^3}$

sorry i dont know hwo to do a fraction in a fraction so 0.33333 = one third
Are you trying to integrate $\displaystyle \frac{4}{x^2}$ i.e.

$\displaystyle \int \frac{4}{x^2}\, dx$?

3. Originally Posted by danny arrigo
Are you trying to integrate $\displaystyle \frac{4}{x^2}$ i.e.

$\displaystyle \int \frac{4}{x^2}\, dx$?
heres the question

A is the point (2,1) on the curve $\displaystyle y=\frac{4}{x^2}$

use calculas to find the gradient of the curve at A

4. Originally Posted by coyoteflare
heres the question

A is the point (2,1) on the curve $\displaystyle y=\frac{4}{x^2}$

use calculas to find the gradient of the curve at A
Oh, you want the slope of the tangent. In this case you'll need the derivative. So

$\displaystyle \frac{d}{dx} \left( \frac{4}{x^2} \right) = \frac{d}{dx} \left( 4 x^{-2} \right)$

Once you have this, evaluate it at $\displaystyle x = 2$ and you'll have your answer.