# Thread: implicit differentiation, tangent line help

1. ## implicit differentiation, tangent line help

Find the equation to the tangent and the normal to the curve $sinx siny = \frac{\sqrt{3}}{4}$

at the point ( $\frac{\pi}{3} , \frac{\pi}{6}$ )

I keep getting this question wrong, here's my working;

using the product rule on sinxsiny
$sinx cosy \frac{dy}{dx} + siny cosx = 0$

$\frac{dy}{dx} = -\frac{ siny cosx}{sinx cosy}$

Is this part correct?

2. Originally Posted by Tweety
Find the equation to the tangent and the normal to the curve $sinx siny = \frac{\sqrt{3}}{4}$

at the point ( $\frac{\pi}{3} , \frac{\pi}{6}$ )

I keep getting this question wrong, here's my working;

using the product rule on sinxsiny
$sinx cosy \frac{dy}{dx} + siny cosx = 0$

$\frac{dy}{dx} = -\frac{ siny cosx}{sinx cosy}$

Is this part correct?

Yes, it is. Now just note that $\frac{\cos x\sin y}{\sin x\cos y}=-\frac{\tan y}{\tan x}$ ...

Tonio