find dy/dx if  (a)y= y ln(xy)
(b) y= cos(x+y)

Dear student,

Kindly find the answer to your query:
$$\begin{gathered} \mathrm{y} = \mathrm{y} \ln \left(\mathrm{xy}\right) \\ 1 = \mathrm{lnxy} \\ 1= \mathrm{lnx}+\mathrm{lny} \\ \mathrm{Differentiating}, \mathrm{we} \mathrm{get} \\ 0=\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}\frac{\mathrm{dy}}{\mathrm{dx}} \\ \frac{\mathrm{dy}}{\mathrm{dx}}= \frac{-\mathrm{y}}{\mathrm{x}} \\ \mathrm{y} = \cos \left(\mathrm{x}+\mathrm{y}\right) \\ \mathrm{Differentiating}, \mathrm{we} \mathrm{get} \\ \frac{\mathrm{dy}}{\mathrm{dx}}= -\sin \left(\mathrm{x}+\mathrm{y}\right)\left(1+\frac{\mathrm{dy}}{\mathrm{dx}}\right) \\ \frac{\mathrm{dy}}{\mathrm{dx}}\left[1+\sin \left(\mathrm{x}+\mathrm{y}\right)\right]= -\sin \left(\mathrm{x}+\mathrm{y}\right) \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\sin \left(\mathrm{x}+\mathrm{y}\right)}{\left[1+\sin \left(\mathrm{x}+\mathrm{y}\right)\right]} \end{gathered}$$

Hope these points will clear your doubt about the topic and if you have more doubts please get back to us on our forum.

Regards