Question #faa9f

1 Answer
Oct 5, 2017

dy/dx=((sec^2(x-y))(3+x^2)^2+(y)(2x))/((3+x^2)+(sec^2(x-y))(3+x^2)^2)

Explanation:

tan(x-y)=y/(3+x^2)

Take the derivative of both sides.

d/dx(tan(x-y))=d/dx(y/(3+x^2))

Solve.

d/dx(tan(x-y))=(1-dy/dx)(sec^2(x-y))

d/dx(y/(3+x^2))=((3+x^2)(dy/dx)-(y)(2x))/(3+x^2)^2

Now we know:

(1-dy/dx)(sec^2(x-y))=((3+x^2)(dy/dx)-(y)(2x))/(3+x^2)^2

We can simplify

(1-dy/dx)(sec^2(x-y))(3+x^2)^2=((3+x^2)(dy/dx)-(y)(2x))

((sec^2(x-y))(3+x^2)^2-(dy/dx)(sec^2(x-y))(3+x^2)^2)=((3+x^2)(dy/dx)-(y)(2x))

((sec^2(x-y))(3+x^2)^2+(y)(2x))=((3+x^2)(dy/dx)+(dy/dx)(sec^2(x-y))(3+x^2)^2)

((sec^2(x-y))(3+x^2)^2+(y)(2x))=(dy/dx)((3+x^2)+(sec^2(x-y))(3+x^2)^2)

((sec^2(x-y))(3+x^2)^2+(y)(2x))/((3+x^2)+(sec^2(x-y))(3+x^2)^2)=dy/dx

Depending on how simplified the answer needs to be, this is technically the solution.