Implicit differentiation

• Jan 6th 2010, 02:58 AM
bobred
Implicit differentiation
Hi

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

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

Thanks
• Jan 6th 2010, 03:06 AM
Prove It
Quote:

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

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

Thanks

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

Use the product rule on the LHS

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

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

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

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

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

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

Go back to your original DE...

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

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

Now try to solve for $\frac{dy}{dx}$.
• Jan 6th 2010, 03:27 AM
bobred
Hi

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

Thanks
• Jan 6th 2010, 03:33 AM
Prove It
Quote:

Originally Posted by bobred
Hi

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

Thanks

Distribute, move everything that has $\frac{dy}{dx}$ to one side, factorise, solve for $\frac{dy}{dx}$.