What is the derivative of ln(2+e^x) ?

The derivative of ln(2+e^x) is (e^x)/(2+e^x)

What is the first derivative of ln(2+e^x) ?

The first derivative of ln(2+e^x) is (e^x)/(2+e^x)

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

\frac{e ^{x}}{2 + e ^{x}}

Solution steps

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

Apply the chain rule:

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

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

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

= \frac{1}{2 + e ^{x}} e^{x}

Simplify

\frac{1}{2 + e ^{x}} e^{x} : \frac{e ^{x}}{2 + e ^{x}}

= \frac{e ^{x}}{2 + e ^{x}}

\int_{- 1}^{1} \frac{2 x}{( x + 7 ) ^{2}} d x

\frac{\partial}{\partial z} (2 x yz)

6 x^{2} y^{^{'}} = y^{^{'}} + 3 x e^{- y}

\frac{d}{d x} (- \frac{1}{2} + \frac{7}{5} x^{3})

\frac{d}{d x} ((x^{3} - 2 x^{2} + 3) (7 x^{2} - 4 x))

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