1. ## Implicit differentiation

Hi

We are asked to solve the following by Implicit differentiation, though I have never encountered this before. Here is the equation

$\displaystyle (x+y)\sin(xy)=1$

Thanks

2. Originally Posted by bobred
Hi

We are asked to solve the following by Implicit differentiation, though I have never encountered this before. Here is the equation

$\displaystyle (x+y)\sin(xy)=1$

Thanks
$\displaystyle \frac{d}{dx}[(x + y)\sin{(xy)}] = \frac{d}{dx}(1)$

Use the product rule on the LHS

$\displaystyle (x + y)\,\frac{d}{dx}[\sin{(xy)}] + \sin{(xy)}\,\frac{d}{dx}(x + y) = 0$

To evaluate $\displaystyle \frac{d}{dx}[\sin{(xy)}]$ you need to use the chain rule.

Let $\displaystyle u = xy$ so that the function becomes $\displaystyle \sin{u}$.

$\displaystyle \frac{du}{dx} = y + x\,\frac{dy}{dx}$

$\displaystyle \frac{d}{du}(\sin{u}) = \cos{u} = \cos{(xy)}$.

Therefore $\displaystyle \frac{d}{dx}[\sin{(xy)}] = \cos{(xy)}\left[y + x\,\frac{dy}{dx}\right]$.

Go back to your original DE...

$\displaystyle (x + y)\,\frac{d}{dx}[\sin{(xy)}] + \sin{(xy)}\,\frac{d}{dx}(x + y) = 0$

$\displaystyle (x + y)\cos{(xy)}\left[y + x\,\frac{dy}{dx}\right] + \sin{(xy)}\left[1 + \frac{dy}{dx}\right] = 0$.

Now try to solve for $\displaystyle \frac{dy}{dx}$.

3. Hi

Its probably obvious but I'm not sure how to go about solving for $\displaystyle \frac{dy}{dx}$
Thanks

4. Originally Posted by bobred
Hi

Its probably obvious but I'm not sure how to go about solving for $\displaystyle \frac{dy}{dx}$
Thanks
Distribute, move everything that has $\displaystyle \frac{dy}{dx}$ to one side, factorise, solve for $\displaystyle \frac{dy}{dx}$.