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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

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

\int cos (x) sin (x) d x

x \to - 5 lim (\frac{∣ x + 5 ∣}{x + 5})

y = \frac{4}{x}, y = \frac{4}{x ^{2}}, x = 5

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