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What is the implicit derivative of #1= x^2y-e^y-xy #?

Answer 1

Evelyn Agard

The answer is #=-(2xy-y)/(x^2-e^y-x)#

Let

#f(x,y)=x^2y-e^y-xy-1#

Then,

The partial derivatives are

#(delf)/(delx)=2xy-y#

#(delf)/(dely)=x^2-e^y-x#

Therefore,

#dy/dx=-((delf)/(delx))/((delf)/(dely))#

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

You can calculate by direct differentiation

#2xy+x^2dy/dx-e^ydy/dx-xdy/dx-y=0#

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

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

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Answer 2

Charles Anctil

To find the implicit derivative of the equation (1 = x^2y - e^y - xy), differentiate both sides of the equation with respect to (x), treating (y) as a function of (x) and using the product rule and chain rule where necessary.

The implicit derivative is:

[\frac{dy}{dx} = \frac{-2xy - e^y}{x^2 - e^y - y}]

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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.