1. ## implicit differentiation

I'm really lost on how to get the derivative of this equation,

x*sin(x*y)=x^2+1

I know that the derivative of a sin is cos, but the problem I am having with this equation is getting the derivative of the xy because I don't know what I'm suppose to do with it.

I know that I'm suppose to use the product rule, but do I also need to use a chain rule or something.... confused...

2. Originally Posted by katchat64
I'm really lost on how to get the derivative of this equation,

x*sin(x*y)=x^2+1

I know that the derivative of a sin is cos, but the problem I am having with this equation is getting the derivative of the xy because I don't know what I'm suppose to do with it.

I know that I'm suppose to use the product rule, but do I also need to use a chain rule or something.... confused...
It will help if you divide both sides by $x$.

Then you have

$\sin{(xy)} = x + x^{-1}$

$\frac{d}{dx}[\sin{(xy)}] = \frac{d}{dx}(x + x^{-1})$

$\frac{d}{dx}(xy)\cdot \frac{d}{du}(\sin{u}) = 1 - x^{-2}$, where $u = xy$

$\left(x\frac{dy}{dx} + y\right)\cos{u} = 1 - \frac{1}{x^2}$

$\left(x\frac{dy}{dx} + y\right)\cos{(xy)} = \frac{x^2 - 1}{x^2}$

$x\frac{dy}{dx} + y = \frac{x^2 - 1}{x^2\cos{(xy)}}$

$x\frac{dy}{dx} = \frac{x^2 - 1}{x^2\cos{(xy)}} - y$

$x\frac{dy}{dx} = \frac{x^2 - 1 - x^2y\cos{(xy)}}{\cos{(xy)}}$

$\frac{dy}{dx} = \frac{x^2 - 1 - x^2y\cos{(xy)}}{x\cos{(xy)}}$.