What is the derivative of y=log_{x}(e) ?

The derivative of y=log_{x}(e) is -1/(xln^2(x))

What is the first derivative of y=log_{x}(e) ?

The first derivative of y=log_{x}(e) is -1/(xln^2(x))

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derivative of

y = lo g_{x} (e)

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

Solution steps

\frac{d}{d x} (lo g_{x} (e))

Apply log rule:

lo g_{a} (b) = \frac{ln ( b )}{ln ( a )}

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (1) = 0

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

= ln (e) \frac{0 \cdot ln ( x ) - \frac{1}{x} \cdot 1}{( ln ( x ) ) ^{2}}

Simplify

ln (e) \frac{0 \cdot ln ( x ) - \frac{1}{x} \cdot 1}{ln ^{2} ( x )} : - \frac{1}{x ln ^{2} ( x )}

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

\int (sec (x) tan (x)) d x

x^{3} cos (x)

\int_{0}^{\frac{π}{12}} t sin (4 t) d t

n = 0 \sum \infty \frac{1}{n + 12}

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

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