# differentiate y = Dsin(kx/y) wrt x

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• Aug 19th 2010, 06:44 PM
tydube
differentiate y = Dsin(kx/y) wrt x
Given a function y = Dsin(kx/y) ,where D and k are constants and (kx/y) is in radians
1. what is dy/dx ?
2. how can you express x as a function of y , x=?
3. what is dx/dy ?
• Aug 19th 2010, 08:13 PM
Prove It
You will need to use the chain rule and implicit differentiation...

$\displaystyle y = D\sin{\frac{kx}{y}}$

Let $\displaystyle u = \frac{kx}{y}$ so that $\displaystyle y= D\sin{u}$.

$\displaystyle \frac{dy}{du} = D\cos{u} = D\cos{\left(\frac{kx}{y}\right)}$.

$\displaystyle \frac{du}{dx} = \frac{d}{dx}\left(\frac{kx}{y}\right)$

$\displaystyle = k\,\frac{d}{dx}\left(\frac{x}{y}\right)$

$\displaystyle = k\left(\frac{y\,\frac{d}{dx}(x) - x\,\frac{d}{dx}(y)}{y^2}\right)$

$\displaystyle = \frac{k\left(y - x\,\frac{dy}{dx}\right)}{y^2}$.

So $\displaystyle \frac{dy}{dx} = \frac{dy}{du}\,\frac{du}{dx}$

$\displaystyle \frac{dy}{dx} = \frac{Dk\left(y - x\,\frac{dy}{dx}\right)\cos{\left(\frac{kx}{y}\rig ht)}}{y^2}$

$\displaystyle y^2\,\frac{dy}{dx} = Dky\cos{\left(\frac{kx}{y}\right)} - Dkx\cos{\left(\frac{kx}{y}\right)}\frac{dy}{dx}$

$\displaystyle y^2\,\frac{dy}{dx} + Dkx\cos{\left(\frac{kx}{y}\right)}\frac{dy}{dx} = Dky\cos{\left(\frac{kx}{y}\right)}$

$\displaystyle \frac{dy}{dx}\left[y^2 + Dkx\cos{\left(\frac{kx}{y}\right)}\right] = Dky\cos{\left(\frac{kx}{y}\right)}$

$\displaystyle \frac{dy}{dx} = \frac{Dky\cos{\left(\frac{kx}{y}\right)}}{y^2 + Dkx\cos{\left(\frac{kx}{y}\right)}}$
• Aug 22nd 2010, 11:14 PM
tydube
Thanks
Thanks for the assistance, much appreciated.(Wink)