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How do you implicitly differentiate #-1=-y^2x-2xy-ye^(xy) #?

Answer 1

#(dy)/(dx)=(y^2+2y+y^2e^(xy))/(2xy+2x+e^(xy)+xye^(xy))#

When we implicitly differentiate a function #f(y)#, we first differentiate #f(y)# with respect to #y# and then multiply by #(dy)/(dx)#. This comes from chain rule as #(df)/(dx)=(df)/(dy)xx(dy)/(dx)#.

Now using product rule for differentiating #1=-y^2x-2xy-ye^(xy)#, we get

#0=-(2y(dy)/(dx)xx x+y^2)-2(1xxy+x xx(dy)/(dx))-((dy)/(dx)e^(xy)+yxxe^(xy)(1xxy+x xx(dy)/(dx)))# or

#(2xy(dy)/(dx)+y^2)+2(y+x(dy)/(dx))+((dy)/(dx)e^(xy)+ye^(xy)(y+x(dy)/(dx)))=0# or

#(dy)/(dx)(2xy+2x+e^(xy)+xye^(xy))=(y^2+2y+y^2e^(xy))# or

#(dy)/(dx)=(y^2+2y+y^2e^(xy))/(2xy+2x+e^(xy)+xye^(xy))#

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Answer 2

Luke Alvarez

To implicitly differentiate -1=-y^2x-2xy-ye^(xy) with respect to x, follow these steps:

Differentiate each term with respect to x.
Apply the chain rule where necessary.
Solve for dy/dx.

The result is:

dy/dx = (2xy^2 + y^3 - 2y - y^2e^(xy)) / (x + 2y^2 + xe^(xy))

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.