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

Question

Christopher Allers · Accepted Answer

#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]#

#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

Samantha Adkison · Answer

undefined To implicitly differentiate the given equation $ \csc(x^2+y^2) = e^{-x-y} $ with respect to $ x $, follow these steps:

1. Differentiate both sides of the equation with respect to $ x $.
2. Apply the chain rule and the derivative of the cosecant function.
3. Simplify the result by isolating the derivative of $ y $ with respect to $ x $.

The implicit differentiation of $ \csc(x^2+y^2) = e^{-x-y} $ with respect to $ x $ yields:

$$ -2x\cot(x^2+y^2)\frac{dy}{dx} - 2y\cot(x^2+y^2)\frac{dy}{dx} = -e^{-x-y} $$

Simplify this expression to solve for $ \frac{dy}{dx} $, which represents the derivative of $ y $ with respect to $ x $.