<s>[INST] Differentiation to tangent and normals The equation of the curve is given by 
$ y = \frac{6}{2x^2+1} $ 
Find the equation of the tangent and the normal at the point when
$ x = -1$ 
My workings 
When $x= -1, y= 2 $ 
$\frac{d}{dx} (Y) = \frac{-24}{(2x^2+1)^2} $ 
Sub $x=-1$ to the above expression to find the gradient ...
$\frac{d}{dx} (Y) = \frac{-8}{3} $ 
Now , to find the equation of the tangent , what can I do ? 
Is it $y=mx+c$ ? 
Sub in $x=-1 , y=2$ to find the y intercept and then continue to find the equation of the tangent ? 
Which is
$y=  \frac{-8}{3} x - \frac{2}{3} $ 
But that don't feel right to me .. so im here to ask . 
 [/INST] HINT: write your function in the form $$y=6(2x^2+1)^{-1}$$ then the derivative is given by
$$y'=6(-1)(2x^2+1)^{-2}\cdot 4x=\frac{-24x}{(2x^2+1)^2}$$
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