Question #da527

1 Answer
Nov 19, 2016

I think there is an #=# sign missing but....

Explanation:

Here you have an Implicit Derivation; in this case the function is kind of "hidden" so that you cannot really write it in the normal, explicit: #y=f(x)#.
Our idea is that #y# represents a function of #x# so that when you derive you must include the term #(dy)/(dx)# to indicate the fact the #y# is, in itsef, a function of #x#.
So for example if you have #y^2# this derived will give you:
#2y(dy)/(dx)#

In your case I suspect that probably there should be a #=# sign somewhere such as:
#8xy+16sin(y)=0#
or other (you need the #=# sign to indicate the functional dependence such as: "this depends from that");

but anyway, if you ONLY derive your expression you should get:

#8y+8x(dy)/(dx)+16cos(y)(dy)/(dx)#

Now it ends here...on the other hand if you consider the original as:
#8xy+16sin(y)=0#
you'll get:
#8y+8x(dy)/(dx)+16cos(y)(dy)/(dx)=0#
you can now collect #(dy)/(dx)#, rearrange and write:
#(dy)/(dx)=-(cancel(8)y)/(cancel(8)(x+2cos(y)))=-(y)/((x+2cos(y))#