What is the derivative of (ln(x+1)/(x+1)) ?

The derivative of (ln(x+1)/(x+1)) is (1-ln(x+1))/((x+1)^2)

What is the first derivative of (ln(x+1)/(x+1)) ?

The first derivative of (ln(x+1)/(x+1)) is (1-ln(x+1))/((x+1)^2)

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

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

Solution steps

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

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

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

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

Simplify

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

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

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

\frac{\partial}{\partial x} ((x - 4)^{2} + y^{2})

(2 x + y cos (x y)) d x + x cos (x y) d y = 0

\frac{\partial}{\partial x} (x^{9} y^{4} - x^{8} y^{3})

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

What is the derivative of (ln(x+1)/(x+1)) ?
The derivative of (ln(x+1)/(x+1)) is (1-ln(x+1))/((x+1)^2)
What is the first derivative of (ln(x+1)/(x+1)) ?
The first derivative of (ln(x+1)/(x+1)) is (1-ln(x+1))/((x+1)^2)