What is the derivative of (ln(x+1)/(e^{x+1)}) ?

The derivative of (ln(x+1)/(e^{x+1)}) is (e^{-x-1}(1-ln(1+x)(1+x)))/(x+1)

What is the first derivative of (ln(x+1)/(e^{x+1)}) ?

The first derivative of (ln(x+1)/(e^{x+1)}) is (e^{-x-1}(1-ln(1+x)(1+x)))/(x+1)

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (ln (x + 1)) = \frac{1}{x + 1}

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

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

Simplify

\frac{\frac{1}{x + 1} e ^{x + 1} - e ^{x + 1} ln ( x + 1 )}{( e ^{x + 1} ) ^{2}} : \frac{e ^{- x - 1} ( 1 - ln ( 1 + x ) ( 1 + x ) )}{x + 1}

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

\frac{8 x}{x + 5}

\int (\frac{2 x}{y}) d x

\frac{d}{d x} ((x^{2} ln (x^{(55)}))^{(\frac{1}{3})})

\frac{\partial}{\partial x} (A e^{- b (A, x, v, t) (x - v t)^{2}})

\frac{d}{d x} (\frac{d y}{d x}) + 25 y = 0

What is the derivative of (ln(x+1)/(e^{x+1)}) ?
The derivative of (ln(x+1)/(e^{x+1)}) is (e^{-x-1}(1-ln(1+x)(1+x)))/(x+1)
What is the first derivative of (ln(x+1)/(e^{x+1)}) ?
The first derivative of (ln(x+1)/(e^{x+1)}) is (e^{-x-1}(1-ln(1+x)(1+x)))/(x+1)