How do you take the derivative of #tan(x - y) = y/(1 + x^2) #?

1 Answer

Have a look, but check my maths.

Explanation:

In this case you'll need to derive implicitly considering that here #y# represents a function of #x# (a little bit...difficult to work out!) and we derive also #y#:
so if you want, say, to derive #y# you get #1*(dy)/(dx)#

so we get (using the Chain Rule on the left and the Quotient Rule on the right):

#1/cos^2(x-y)*[1-1(dy)/(dx)]=((dy)/(dx)(1+x^2)-2xy)/(1+x^2)^2#

rearrange to isolate #(dy)/(dx)#:

#((1+x^2)^2)/cos^2(x-y)-((1+x^2)^2)/cos^2(x-y)(dy)/(dx)=(dy)/(dx)(1+x^2)-2xy#

#(dy)/(dx)[(1+x^2)+((1+x^2)^2)/cos^2(x-y)]=2xy+((1+x^2)^2)/cos^2(x-y)#

Multiply the non-fractions by #cos^2(x-y)/cos^2(x-y)# to get common denominators, cancel out those denominators, then divide by the resultant left-side bracketed contents:

#(dy)/(dx)=[2xycos^2(x-y)+(1+x^2)^2]/[(1+x^2)cos^2(x-y)+(1+x^2)^2]#