What is the derivative of (1-ln(x)/(4x^2)) ?

The derivative of (1-ln(x)/(4x^2)) is -(-2ln(x)+3)/(4x^3)

What is the first derivative of (1-ln(x)/(4x^2)) ?

The first derivative of (1-ln(x)/(4x^2)) is -(-2ln(x)+3)/(4x^3)

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

- \frac{- 2 ln ( x ) + 3}{4 x ^{3}}

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

= \frac{1}{4} \cdot \frac{( - \frac{1}{x} ) x ^{2} - 2 x ( 1 - ln ( x ) )}{( x ^{2} ) ^{2}}

Simplify

\frac{1}{4} \cdot \frac{( - \frac{1}{x} ) x ^{2} - 2 x ( 1 - ln ( x ) )}{( x ^{2} ) ^{2}} : - \frac{- 2 ln ( x ) + 3}{4 x ^{3}}

= - \frac{- 2 ln ( x ) + 3}{4 x ^{3}}

\int \frac{cos ( 5 x )}{1 + sin ^{2} ( 5 x )} d x

\frac{9}{s ( s + 1 ) ( s + 9 )}

\int \frac{x ^{2} - 1}{x ^{2} + 1} d x

y^{^{''}} - 16 y^{^{'}} + 100 y = 0

\frac{\partial}{\partial y} (4 x y^{3} (z^{2} + 1)^{3})

What is the derivative of (1-ln(x)/(4x^2)) ?
The derivative of (1-ln(x)/(4x^2)) is -(-2ln(x)+3)/(4x^3)
What is the first derivative of (1-ln(x)/(4x^2)) ?
The first derivative of (1-ln(x)/(4x^2)) is -(-2ln(x)+3)/(4x^3)