# Thread: equations help ( differentiation )

1. ## equations help ( differentiation )

The curve $y= cx + \frac{d}{x}$ has gradient 6 at the point $( \frac{1}{2} , 1 )$ find c and d

my working,

sub in the values for x and y and solve simultaneously but i dont get the right answer.....

first equation = $\frac{1}{2}c + \frac{d}{\frac{1}{2}}$ which I simplfied to $2c + 2d = 1$ is this simplification correct?

$\frac{dy}{dx} = c -dx^{-2}= 6$

these are my two equations, and when I try to solve for 'c' and 'd' i dont get the right answer?

2. Hello,
Originally Posted by Tweety
The curve $y= cx + \frac{d}{x}$ has gradient 6 at the point $( \frac{1}{2} , 1 )$ find c and d

my working,

sub in the values for x and y and solve simultaneously but i dont get the right answer.....

first equation = $\frac{1}{2}c + \frac{d}{\frac{1}{2}}$ which I simplfied to $2c + 2d = 1$ is this simplification correct?
It's $1=\frac 12 c +\frac{d}{\frac 12}$
There's a problem with your simplification :
$\frac{d}{\frac 12}=2d$
so you have $1=\frac 12 c+2d$

$\frac{dy}{dx} = c -dx^{-2}= 6$

these are my two equations, and when I try to solve for 'c' and 'd' i dont get the right answer?
This one is correct, though it is an equation for x=1/2

3. how do i evaluate $-dx^{-2}$ when x = $\frac{1}{2}$

is this method correct..

$\frac{1}{2}^{-2}$ = $-\frac{1}{2} \times -\frac{1}{2} = \frac{1}{4}$

so the equation is $c -\frac{1}{4}d = 6$ ?

4. Originally Posted by Tweety
how do i evaluate $-dx^{-2}$ when x = $\frac{1}{2}$

is this method correct..

$\frac{1}{2}^{-2}$ = $-\frac{1}{2} \times -\frac{1}{2} = \frac{1}{4}$

so the equation is $c -\frac{1}{4}d = 6$ ?
wow... I think you've got some problems in algebra :s

$-x^{-2}=-\frac{1}{x^2}$

So when x=1/2, we have $x^2=\left(\frac 12\right)^2=\frac{1^2}{2^2}=\frac 14$

---> $\frac{1}{x^2}=\frac{1}{\frac 14}=4$

-----> $-dx^{-2}=-4d$

5. Originally Posted by Moo
wow... I think you've got some problems in algebra :s

$-x^{-2}=-\frac{1}{x^2}$

So when x=1/2, we have $x^2=\left(\frac 12\right)^2=\frac{1^2}{2^2}=\frac 14$

---> $\frac{1}{x^2}=\frac{1}{\frac 14}=4$

-----> $-dx^{-2}=-4d$
Yes thats why I ask for help !!

thanks anyways..........