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

The derivative of sin((2x/(x^4-4x))) is -(6x^2cos(2/(x^3-4)))/((x^3-4)^2)

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

The first derivative of sin((2x/(x^4-4x))) is -(6x^2cos(2/(x^3-4)))/((x^3-4)^2)

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

- \frac{6 x ^{2} cos ( \frac{2}{x ^{3} - 4} )}{( x ^{3} - 4 ) ^{2}}

Solution steps

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

Simplify

sin (\frac{2 x}{x ^{4} - 4 x}) : sin (\frac{2}{x ^{3} - 4})

= \frac{d}{d x} (sin (\frac{2}{x ^{3} - 4}))

Apply the chain rule:

cos (\frac{2}{x ^{3} - 4}) \frac{d}{d x} (\frac{2}{x ^{3} - 4})

= cos (\frac{2}{x ^{3} - 4}) \frac{d}{d x} (\frac{2}{x ^{3} - 4})

\frac{d}{d x} (\frac{2}{x ^{3} - 4}) = - \frac{6 x ^{2}}{( x ^{3} - 4 ) ^{2}}

= cos (\frac{2}{x ^{3} - 4}) (- \frac{6 x ^{2}}{( x ^{3} - 4 ) ^{2}})

Simplify

cos (\frac{2}{x ^{3} - 4}) (- \frac{6 x ^{2}}{( x ^{3} - 4 ) ^{2}}) : - \frac{6 x ^{2} cos ( \frac{2}{x ^{3} - 4} )}{( x ^{3} - 4 ) ^{2}}

= - \frac{6 x ^{2} cos ( \frac{2}{x ^{3} - 4} )}{( x ^{3} - 4 ) ^{2}}

n = 0 \sum \infty \frac{9}{n ( n + 3 )}

\int (x^{2}) d x

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

\frac{d}{d x} (4 x^{3} sec (7 x))

(x^{2} + 1) y^{^{'}} + 7 x (y - 1) = 0

What is the derivative of sin((2x/(x^4-4x))) ?
The derivative of sin((2x/(x^4-4x))) is -(6x^2cos(2/(x^3-4)))/((x^3-4)^2)
What is the first derivative of sin((2x/(x^4-4x))) ?
The first derivative of sin((2x/(x^4-4x))) is -(6x^2cos(2/(x^3-4)))/((x^3-4)^2)