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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derivative of

\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 )}

\int sin^{2} (5 θ) d θ

\int \frac{x ^{2} - 5 x + 3}{x - 7} d x

x^{2} + y^{2} = x y y^{^{'}}

e^{- 2 s} (\frac{1}{4 s})

\int x (4 - x^{2})^{\frac{1}{2}} d x

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))