What is the derivative of-y/(x+e^y) ?

The derivative of-y/(x+e^y) is -(x+e^y-e^yy)/((x+e^y)^2)

What is the first derivative of-y/(x+e^y) ?

The first derivative of-y/(x+e^y) is -(x+e^y-e^yy)/((x+e^y)^2)

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derivative of

- \frac{y}{x + e ^{y}}

- \frac{x + e ^{y} - e ^{y} y}{( x + e ^{y} ) ^{2}}

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= - \frac{\frac{d y}{d y} ( x + e ^{y} ) - \frac{d}{d y} ( x + e ^{y} ) y}{( x + e ^{y} ) ^{2}}

\frac{d y}{d y} = 1

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

= - \frac{1 \cdot ( x + e ^{y} ) - e ^{y} y}{( x + e ^{y} ) ^{2}}

Simplify

= - \frac{x + e ^{y} - e ^{y} y}{( x + e ^{y} ) ^{2}}

\int_{0}^{3} \frac{6 t}{( t - 4 ) ^{2}} d t

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

\frac{\partial}{\partial x} (y - x^{2} y e^{- y})

sin (x^{2}), 0

y^{^{''}} + y^{^{'}} - 2 y = sin (x)

What is the derivative of-y/(x+e^y) ?
The derivative of-y/(x+e^y) is -(x+e^y-e^yy)/((x+e^y)^2)
What is the first derivative of-y/(x+e^y) ?
The first derivative of-y/(x+e^y) is -(x+e^y-e^yy)/((x+e^y)^2)