What is the tangent of xy+1+e=e^{-xy},(1,-1) ?

The tangent of xy+1+e=e^{-xy},(1,-1) is y=(-\W_{0}(e^{e+1})+1+e)x+\W_{0}(e^{e+1})-2-e

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tangent of xy+1+e=e^{-xy},(1,-1)

tangent of

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

y = (- W_{0} (e^{e + 1}) + 1 + e) x + W_{0} (e^{e + 1}) - 2 - e

Solution steps

Find the slope of

x y + 1 + e = e^{- x y} : \frac{d y}{d x} = - \frac{W _{0} ( e ^{e + 1} ) - 1 - e}{x ^{2}}

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 = - W_{0} (e^{e + 1}) + 1 + e

Find the line with slope m=

- W_{0} (e^{e + 1}) + 1 + e

and passing through

(1, - 1) : y = (- W_{0} (e^{e + 1}) + 1 + e) x + W_{0} (e^{e + 1}) - 2 - e

y = (- W_{0} (e^{e + 1}) + 1 + e) x + W_{0} (e^{e + 1}) - 2 - e

\int 6 x y^{2} d y

f (x) = e^{x}, at x = ln (14)

x \to 2 lim (2 x - 4 + 8)

\int (cot (x)) d x

\frac{d}{d x} (\frac{8}{3 x ^{2}} + \frac{3}{2 x} + 2)

What is the tangent of xy+1+e=e^{-xy},(1,-1) ?
The tangent of xy+1+e=e^{-xy},(1,-1) is y=(-\W_{0}(e^{e+1})+1+e)x+\W_{0}(e^{e+1})-2-e