# How do you find #(dy)/(dx)# given #ln(xy)=cos(y^4)#?

##### 1 Answer

Jun 23, 2017

#### Explanation:

Differentiate as normal, but remember that differentiating any function of

First, we can simplify the natural logarithm function using

#lnx+lny=cos(y^4)#

Then, differentiating:

#1/x+1/ydy/dx=-4y^3sin(y^4)dy/dx#

Group the

#1/ydy/dx+4y^3sin(y^4)dy/dx=-1/x#

#dy/dx(1/y+4y^3sin(y^4))=-1/x#

#dy/dx=(-1/x)/(1/y+4y^3sin(y^4))#

Multiplying the fraction by

#dy/dx=(-y)/(x+4xy^4sin(y^4))#