Implicit differentiation

**bobred** · January 6th 2010, 01:58 AM

Hi

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

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

Can anyone please help and explain what they are doing?

Thanks

**Prove It** · January 6th 2010, 02:06 AM

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

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

Can anyone please help and explain what they are doing?

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 $\frac{dy}{dx}$ .

**bobred** · January 6th 2010, 02:27 AM

Hi

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

**Prove It** · January 6th 2010, 02:33 AM

Originally Posted by bobred

Hi

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

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

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