What is the tangent of ln(xy)+y^2=2y,(e,1) ?

The tangent of ln(xy)+y^2=2y,(e,1) is y=-1/e x+2

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tangent of ln(xy)+y^2=2y,(e,1)

tangent of

ln (x y) + y^{2} = 2 y, (e, 1)

y = - \frac{1}{e} x + 2

Solution steps

Find the slope of

ln (x y) + y^{2} = 2 y : \frac{d y}{d x} = - \frac{y}{x ( 1 + 2 y ^{2} - 2 y )}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - \frac{1}{e}

Find the line with slope m=

- \frac{1}{e}

and passing through

(e, 1) : y = - \frac{1}{e} x + 2

y = - \frac{1}{e} x + 2

What is the tangent of ln(xy)+y^2=2y,(e,1) ?
The tangent of ln(xy)+y^2=2y,(e,1) is y=-1/e x+2