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 )}

Compute the slope of

ln (x y) + y^{2} = 2 y

(e, 1) : 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

\frac{\partial}{\partial x} (y^{4} \frac{1}{x})

\int x sin (1 + x^{\frac{3}{2}}) d x

\int sin (6 π x) d x

\frac{\partial}{\partial x} (ln (\frac{2}{x}))

x^{2} y^{^{''}} + x y^{^{'}} - y = 2 x

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