What is the derivative of 1/(2xsqrt(ln(x+1))) ?

The derivative of 1/(2xsqrt(ln(x+1))) is -(2ln(x)+3)/(4x^2(ln(x)+1)sqrt(ln(x)+1))

What is the first derivative of 1/(2xsqrt(ln(x+1))) ?

The first derivative of 1/(2xsqrt(ln(x+1))) is -(2ln(x)+3)/(4x^2(ln(x)+1)sqrt(ln(x)+1))

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

- \frac{2 ln ( x ) + 3}{4 x ^{2} ( ln ( x ) + 1 ) ln ( x ) + 1}

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

\frac{d}{d x} (x ln (x) + 1) = \frac{2 ln ( x ) + 3}{2 ln ( x ) + 1}

= \frac{1}{2} - \frac{1}{( x ln ( x ) + 1 ) ^{2}} \cdot \frac{2 ln ( x ) + 3}{2 ln ( x ) + 1}

Simplify

\frac{1}{2} - \frac{1}{( x ln ( x ) + 1 ) ^{2}} \cdot \frac{2 ln ( x ) + 3}{2 ln ( x ) + 1} : - \frac{2 ln ( x ) + 3}{4 x ^{2} ( ln ( x ) + 1 ) ln ( x ) + 1}

= - \frac{2 ln ( x ) + 3}{4 x ^{2} ( ln ( x ) + 1 ) ln ( x ) + 1}

\int cot^{3} (x) \cdot csc^{3} (x) d x

n = 1 \sum \infty \frac{2 n !}{n ^{n}}

r (2 r + 1)

\frac{d y}{d t} = 5 (y - t^{2})

\int \frac{x ^{3}}{( 2 + x ^{4} ) ^{9}} d x

What is the derivative of 1/(2xsqrt(ln(x+1))) ?
The derivative of 1/(2xsqrt(ln(x+1))) is -(2ln(x)+3)/(4x^2(ln(x)+1)sqrt(ln(x)+1))
What is the first derivative of 1/(2xsqrt(ln(x+1))) ?
The first derivative of 1/(2xsqrt(ln(x+1))) is -(2ln(x)+3)/(4x^2(ln(x)+1)sqrt(ln(x)+1))