What is the derivative of (1-log_{e}(x))/(x^2) ?

The derivative of (1-log_{e}(x))/(x^2) is -(-2ln(x)+3)/(x^3)

What is the first derivative of (1-log_{e}(x))/(x^2) ?

The first derivative of (1-log_{e}(x))/(x^2) is -(-2ln(x)+3)/(x^3)

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derivative of

\frac{1 - lo g _{e} ( x )}{x ^{2}}

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

Solution steps

\frac{d}{d x} (\frac{1 - lo g _{e} ( x )}{x ^{2}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 1 - lo g _{e} ( x ) ) x ^{2} - \frac{d}{d x} ( x ^{2} ) ( 1 - lo g _{e} ( x ) )}{( x ^{2} ) ^{2}}

\frac{d}{d x} (1 - lo g_{e} (x)) = - \frac{1}{x}

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

= \frac{( - \frac{1}{x} ) x ^{2} - 2 x ( 1 - lo g _{e} ( x ) )}{( x ^{2} ) ^{2}}

Simplify

\frac{( - \frac{1}{x} ) x ^{2} - 2 x ( 1 - lo g _{e} ( x ) )}{( x ^{2} ) ^{2}} : - \frac{- 2 ln ( x ) + 3}{x ^{3}}

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

\int sec^{2} (θ) d θ

f (x) = 7 x^{4} + 4 x^{2} x 3 + 1

\int (1 + x) d x

\int \frac{1}{ln ( n )}

\int_{0}^{1} x \cdot ln (x) d x

What is the derivative of (1-log_{e}(x))/(x^2) ?
The derivative of (1-log_{e}(x))/(x^2) is -(-2ln(x)+3)/(x^3)
What is the first derivative of (1-log_{e}(x))/(x^2) ?
The first derivative of (1-log_{e}(x))/(x^2) is -(-2ln(x)+3)/(x^3)