What is the derivative of y=(csc(4x))/(ln(x)) ?

The derivative of y=(csc(4x))/(ln(x)) is (-4xcot(4x)csc(4x)ln(x)-csc(4x))/(xln^2(x))

What is the first derivative of y=(csc(4x))/(ln(x)) ?

The first derivative of y=(csc(4x))/(ln(x)) is (-4xcot(4x)csc(4x)ln(x)-csc(4x))/(xln^2(x))

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derivative of

y = \frac{csc ( 4 x )}{ln ( x )}

\frac{- 4 x cot ( 4 x ) csc ( 4 x ) ln ( x ) - csc ( 4 x )}{x ln ^{2} ( x )}

Solution steps

\frac{d}{d x} (\frac{csc ( 4 x )}{ln ( x )})

Apply the Quotient Rule:

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

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

\frac{d}{d x} (csc (4 x)) = - cot (4 x) csc (4 x) \cdot 4

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

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

Simplify

\frac{( - cot ( 4 x ) csc ( 4 x ) \cdot 4 ) ln ( x ) - \frac{1}{x} csc ( 4 x )}{ln ^{2} ( x )} : \frac{- 4 x cot ( 4 x ) csc ( 4 x ) ln ( x ) - csc ( 4 x )}{x ln ^{2} ( x )}

= \frac{- 4 x cot ( 4 x ) csc ( 4 x ) ln ( x ) - csc ( 4 x )}{x ln ^{2} ( x )}

x^{2} \frac{d ^{2} y}{d x ^{2}} + 4 x \frac{d y}{d x} + 2 y = 8 x

y = x^{2} - 8

\int_{- \infty}^{1} x e^{2 x} d x

\int x^{2} - a^{2} d x

- \frac{1}{4 s}

What is the derivative of y=(csc(4x))/(ln(x)) ?
The derivative of y=(csc(4x))/(ln(x)) is (-4xcot(4x)csc(4x)ln(x)-csc(4x))/(xln^2(x))
What is the first derivative of y=(csc(4x))/(ln(x)) ?
The first derivative of y=(csc(4x))/(ln(x)) is (-4xcot(4x)csc(4x)ln(x)-csc(4x))/(xln^2(x))