What is the d/(dy)((x)(1-e^{-x}-e^{-y}+e^{-x-y})) ?

The d/(dy)((x)(1-e^{-x}-e^{-y}+e^{-x-y})) is x(-e^{-x-y}+e^{-y})

What is the first d/(dy)((x)(1-e^{-x}-e^{-y}+e^{-x-y})) ?

The first d/(dy)((x)(1-e^{-x}-e^{-y}+e^{-x-y})) is x(-e^{-x-y}+e^{-y})

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d y} ((x) (1 - e^{- x} - e^{- y} + e^{- x - y}))

x (- e^{- x - y} + e^{- y})

Solution steps

\frac{d}{d y} ((x) (1 - e^{- x} - e^{- y} + e^{- x - y}))

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= x \frac{d}{d y} (1 - e^{- x} - e^{- y} + e^{- x - y})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= x (\frac{d}{d y} (1) - \frac{d}{d y} (e^{- x}) - \frac{d}{d y} (e^{- y}) + \frac{d}{d y} (e^{- x - y}))

\frac{d}{d y} (1) = 0

\frac{d}{d y} (e^{- x}) = 0

\frac{d}{d y} (e^{- y}) = - e^{- y}

\frac{d}{d y} (e^{- x - y}) = - e^{- x - y}

= x (0 - 0 - (- e^{- y}) - e^{- x - y})

Simplify

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

(t (2 - t)) \cdot 2

(- u, v, u cos (v), 0) \cdot (cos (v), v, - 2 u)

U (t - 4) (2 - t^{2}) + 2 e^{π t} + sin (3 t) - t^{3}

I n (x \cdot y \cdot z^{4})

\frac{9 x ^{2} - 4}{3 x - 2}

What is the d/(dy)((x)(1-e^{-x}-e^{-y}+e^{-x-y})) ?
The d/(dy)((x)(1-e^{-x}-e^{-y}+e^{-x-y})) is x(-e^{-x-y}+e^{-y})
What is the first d/(dy)((x)(1-e^{-x}-e^{-y}+e^{-x-y})) ?
The first d/(dy)((x)(1-e^{-x}-e^{-y}+e^{-x-y})) is x(-e^{-x-y}+e^{-y})