# Math Help - Help with differentiation..

1. ## Help with differentiation..

Show that the equation of the tangent to y = 1 / x at the point for which x = p is p^2y + x = 2p. At what point on the curve is the equation of the tangent 9y + x + 6 = 0?

Differentiating:
$f'(x) = -1/x^2$

$f'(p) = -1 / p^2$

Shouldn't the equation be worked as follow? It's not working..
$y - (1 / p) = -p^-2(x - p)$

2. Originally Posted by struck
Show that the equation of the tangent to y = 1 / x at the point for which x = p is p^2y + x = 2p. At what point on the curve is the equation of the tangent 9y + x + 6 = 0?

Differentiating:
$f'(x) = -1/x^2$

$f'(p) = -1 / p^2$
$y - (1 / p) = -p^-2(x - p)$
$y - \dfrac1p = -\dfrac1{p^2}(x - p)$ Multiply the equation by $p^2$
$p^2 y -p = -x+p$ Rearrange the equation:
$p^2y + x = 2p$