What is the derivative of f(x)= x/(ln(x)) ?

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

What is the first derivative of f(x)= x/(ln(x)) ?

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

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derivative of

f (x) = \frac{x}{ln ( x )}

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

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d x}{d x} = 1

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

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

Simplify

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

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

y^{^{''}} + 6 y^{^{'}} + 19 y = 0

y^{^{''}} = 0

\int \frac{1}{25 + ( x - 6 ) ^{2}} d x

\frac{1}{5 s ^{2} + 0.2 s}

\frac{\partial}{\partial y} (e^{5 y} cos (- 5 x))

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