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