# How do you use implicit differentiation to find dy/dx given #cos2xsec3y+1=0#?

##### 1 Answer

#### Explanation:

Remember the product rule:

#d/dxf(x)g(x) = f(x)g'(x) + g(x)f'(x)#

In this case we differentiate like this:

#d/dx(cos(2x)sec(3y)+1) = d/dx0#

#d/dxcos(2x)sec(3y) = 0#

Let's say that

#f'(x) = d/dxcos(2x) = -sin(2x)*d/dx(2x) = -2sin(2x)#

#g'(x) = d/dxsec(3y) = sec(3y)tan(3y) * d/dx3y = 3(dy/dx)sec(3y)tan(3y)#

Therefore we can differentiate our equation like this:

#d/dxcos(2x)sec(3y) = 0#

#cos(2x)(3dy/dxsec(3y)tan(3y))+sec(3y)(-2sin(2x))=0#

All that is left to do is solve for

#3cos(2x)sec(3y)tan(3y)dy/dx - 2sec(3y)sin(2x) = 0#

#3cos(2x)sec(3y)tan(3y)dy/dx = 2sec(3y)sin(2x)#

#dy/dx = (2sec(3y)sin(2x))/(3cos(2x)sec(3y)tan(3y)#

#dy/dx = (2color(blue)cancelcolor(black)sec(3y)sin(2x))/(3color(red)cancelcolor(black)cos(2x)color(blue)cancelcolor(black)sec(3y)tan(3y)) * (1/cos(2x))/color(red)cancelcolor(black)(1/cos(2x))#

#dy/dx = (2tan(2x))/(3tan(3y))#

*Final Answer*