What is the derivative of g(x)=(ln(x))/(x^2) ?

The derivative of g(x)=(ln(x))/(x^2) is (1-2ln(x))/(x^3)

What is the first derivative of g(x)=(ln(x))/(x^2) ?

The first derivative of g(x)=(ln(x))/(x^2) is (1-2ln(x))/(x^3)

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derivative of

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

\frac{1 - 2 ln ( x )}{x ^{3}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

= \frac{1 - 2 ln ( x )}{x ^{3}}

\frac{d}{d x} ((ln (x))^{13})

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

6 x^{2} - 4 x + 5

\int \frac{( x ^{4} - 5 x ^{2} + 3 x - 4 )}{x} d x

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

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