Hi

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

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

Can anyone please help and explain what they are doing?

Thanks

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- Jan 6th 2010, 01:58 AMbobredImplicit differentiation
Hi

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

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

Can anyone please help and explain what they are doing?

Thanks - Jan 6th 2010, 02:06 AMProve It
$\displaystyle \frac{d}{dx}[(x + y)\sin{(xy)}] = \frac{d}{dx}(1)$

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 $\displaystyle \frac{dy}{dx}$. - Jan 6th 2010, 02:27 AMbobred
Hi

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

$

Thanks

- Jan 6th 2010, 02:33 AMProve It