How do you find the derivative of #xe^y=y-1#?

Use implicit differentiation. On the left hand side use the product rule.

We are differentiating with respect to x

Factor left hand side

To find the derivative of ( xe^y = y - 1 ) with respect to ( x ), you need to use implicit differentiation. Here's the step-by-step process:

1. Take the derivative of both sides of the equation with respect to ( x ).
2. Apply the product rule on the left side where ( xe^y ) is considered as the product of two functions, ( x ) and ( e^y ).
3. Apply the chain rule to differentiate ( e^y ) with respect to ( x ), considering ( y ) as a function of ( x ).
4. Solve for ( \frac{dy}{dx} ), which represents the derivative of ( y ) with respect to ( x ).

Starting with ( xe^y = y - 1 ):

1. Differentiating both sides with respect to ( x ): [ \frac{d}{dx}(xe^y) = \frac{d}{dx}(y - 1) ]

2. Applying the product rule on the left side: [ e^y + xe^y\frac{dy}{dx} = \frac{dy}{dx} ]

3. Differentiating ( e^y ) with respect to ( x ) using the chain rule: [ e^y\frac{dy}{dx} + xe^y\frac{dy}{dx} = \frac{dy}{dx} ]

4. Solving for ( \frac{dy}{dx} ): [ \frac{dy}{dx}(e^y + xe^y) = \frac{dy}{dx} ] [ \frac{dy}{dx}(1 + x)e^y = \frac{dy}{dx} ]

[ (1 + x)e^y = 1 ]

[ e^y = \frac{1}{1 + x} ]

[ y = \ln{\left(\frac{1}{1 + x}\right)} ]

Now, knowing ( y ), you can differentiate implicitly with respect to ( x ) to find ( \frac{dy}{dx} ): [ \frac{d}{dx}\left(\ln{\left(\frac{1}{1 + x}\right)}\right) = \frac{d}{dx}(y) ]

[ \frac{d}{dx}\left(\ln{\left(\frac{1}{1 + x}\right)}\right) = \frac{d}{dx}(y) ] [ \frac{d}{dx}\left(\ln{\left(\frac{1}{1 + x}\right)}\right) = \frac{d}{dx}(y) ] [ \frac{d}{dx}\left(-\ln{(1 + x)}\right) = \frac{dy}{dx} ] [ -\frac{1}{1 + x} = \frac{dy}{dx} ]

So, the derivative of ( xe^y = y - 1 ) with respect to ( x ) is ( -\frac{1}{1 + x} ).