Question

How do you implicitly differentiate `csc(x^2+y^2)=e^-x-y`?

Answer 1

`y'=(-2x^2*csc (x^2+y^2)*cot (x^2+y^2)+e^(-x))/[2y*csc (x^2+y^2)*cot (x^2+y^2)-1]`

Explanation:

`csc (x^2+y^2)=e^(-x)-y`

differentiate both sides of the equation with respect to x

`d/dx(csc (x^2+y^2))=d/dx(e^(-x)-y)`

`-csc (x^2+y^2)*cot (x^2+y^2)*d/dx(x^2+y^2)=d/dx(e^(-x))-d/dx(y)`

`-csc (x^2+y^2)*cot (x^2+y^2)(2x^2+2y*y')=(e^(-x))*(-1)-y'`

`-csc (x^2+y^2)*cot (x^2+y^2)(2x^2+2y*y')=-e^(-x)-y'`

Multiplication then transposition

`(2x^2(-csc (x^2+y^2)*cot (x^2+y^2))+e^(-x)=`
`2y*y'*(csc (x^2+y^2)*cot (x^2+y^2))-y'`

factor out the y'

`-2x^2csc (x^2+y^2)*cot (x^2+y^2)+e^(-x)=`
`[2y*csc (x^2+y^2)*cot (x^2+y^2)-1]y'`

Divide both sides by `[2y*csc (x^2+y^2)*cot (x^2+y^2)-1]`

`y'=(-2x^2*csc (x^2+y^2)*cot (x^2+y^2)+e^(-x))/[2y*csc (x^2+y^2)*cot (x^2+y^2)-1]`

God bless....I hope the explanation is useful