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

Question

Charles Arocha · Accepted Answer

#dy/dx=(2xcsc(x^2/y^2)cot(x^2/y^2)+y^3e^(xy))/(xy(2xcsc(x^2/y^2)cot(x^2/y^2)-ye^(xy))#

Implicit differentiation is no different from explicit differentiation. Just remember that differentiating a function of #y# causes the chain rule to be in effect. Recall also that #d/dxcscx=-cscxcotx# and #d/dxe^x=e^x#.

#csc(x^2/y^2)=e^(xy)#

#d/dxcsc(x^2/y^2)=d/dxe^(xy)#

#-csc(x^2/y^2)cot(x^2/y^2)*d/dx(x^2y^-2)=e^(xy)*d/dx(xy)#

Use the product rule to find these derivatives. Recall that while #d/dxx^2=2x#, #d/dxy^2=2ydy/dx#.

#-csc(x^2/y^2)cot(x^2/y^2)(2xy^-2-2x^2y^-1dy/dx)=e^(xy)(y+xdy/dx)#

Expanding and rearranging to group #dy/dx# terms:

#((2x^2csc(x^2/y^2)cot(x^2/y^2))/y-xe^(xy))dy/dx=(2xcsc(x^2/y^2)cot(x^2/y^2))/y^2+ye^(xy)#

Common denominators:

#((2x^2csc(x^2/y^2)cot(x^2/y^2)-xye^(xy))/y)dy/dx=(2xcsc(x^2/y^2)cot(x^2/y^2)+y^3e^(xy))/y^2#

Solving for #dy/dx#:

#dy/dx=(2xcsc(x^2/y^2)cot(x^2/y^2)+y^3e^(xy))/y^2*y/(2x^2csc(x^2/y^2)cot(x^2/y^2)-xye^(xy))#

#dy/dx=(2xcsc(x^2/y^2)cot(x^2/y^2)+y^3e^(xy))/(xy(2xcsc(x^2/y^2)cot(x^2/y^2)-ye^(xy))#

Christopher Agresta · Answer

undefined To implicitly differentiate $ \csc(x^2/y^2) = e^{xy} $, first, apply the chain rule and the derivative of $ \csc(u) $, where $ u = x^2/y^2 $. Then differentiate $ e^{xy} $ with respect to $ x $.

The result after implicit differentiation is:

$$ \frac{-2x \cot(x^2/y^2)}{y^2} - y^2 e^{xy} = 0 $$