What is the derivative of (9x/(ln(x))) ?

The derivative of (9x/(ln(x))) is (9(ln(x)-1))/(ln^2(x))

What is the first derivative of (9x/(ln(x))) ?

The first derivative of (9x/(ln(x))) is (9(ln(x)-1))/(ln^2(x))

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\frac{d}{d x} (\frac{9 x}{ln ( x )})

\frac{9 ( ln ( x ) - 1 )}{ln ^{2} ( x )}

Solution steps

\frac{d}{d x} (\frac{9 x}{ln ( x )})

Take the constant out:

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

= 9 \frac{d}{d x} (\frac{x}{ln ( x )})

Apply the Quotient Rule:

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

= 9 \cdot \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}

= 9 \cdot \frac{1 \cdot ln ( x ) - \frac{1}{x} x}{( ln ( x ) ) ^{2}}

Simplify

9 \cdot \frac{1 \cdot ln ( x ) - \frac{1}{x} x}{ln ^{2} ( x )} : \frac{9 ( ln ( x ) - 1 )}{ln ^{2} ( x )}

= \frac{9 ( ln ( x ) - 1 )}{ln ^{2} ( x )}

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

(x - \frac{6}{x})^{8}

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

2 sec (x) + 5 cot (x)

\int_{0}^{π} 2 cos^{2} (\frac{x}{2}) d x

What is the derivative of (9x/(ln(x))) ?
The derivative of (9x/(ln(x))) is (9(ln(x)-1))/(ln^2(x))
What is the first derivative of (9x/(ln(x))) ?
The first derivative of (9x/(ln(x))) is (9(ln(x)-1))/(ln^2(x))