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

The derivative of (ln(x-1)/(ln^2(x))) is (-ln(x)+2)/(xln^3(x))

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{\frac{1}{x} ln ^{2} ( x ) - \frac{2 l n ( x )}{x} ( ln ( x ) - 1 )}{( ln ^{2} ( x ) ) ^{2}} : \frac{- ln ( x ) + 2}{x ln ^{3} ( x )}

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

\frac{12}{1 + x ^{4}}, 6 x^{2}

2 x^{x}

n = 0 \sum \infty \frac{2 ^{2}}{3 ^{2}}

\frac{\partial}{\partial s} (- 3 s^{2} t^{2})

\frac{d y}{d x} = \frac{y ^{3}}{3 x + 1}, y (0) = - 1

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