The equation of tangent to curve $y = e^{- x}$ at the points where the curve cuts the line x = 1 is

x+ y = e

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e (x + y) = 1

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y + ex = 1

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ey+x=2

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Solution

The correct option is D ey+x=2
$y = e^{- x} (x = 1) (y = \frac{1}{e}) \frac{d y}{d x} = - e^{- x} = - \frac{1}{e} (y - \frac{1}{e}) = \frac{- (x - 1)}{e} e y + x = 2$

Suggest Corrections

Similar questions

y = e^{- | x |}

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y = e^{- | x |}

y = e^{- 2 | x |}

x = \frac{1}{2}

y = (x^{3} - 1) (x - 2)

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