What is the derivative of 2/3 sin((x+1/(x-1))) ?

The derivative of 2/3 sin((x+1/(x-1))) is -(4cos(\frac{x+1)/(x-1))}{3(x-1)^2}

What is the first derivative of 2/3 sin((x+1/(x-1))) ?

The first derivative of 2/3 sin((x+1/(x-1))) is -(4cos(\frac{x+1)/(x-1))}{3(x-1)^2}

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\frac{d}{d x} (\frac{2}{3} sin (\frac{x + 1}{x - 1}))

- \frac{4 cos ( \frac{x + 1}{x - 1} )}{3 ( x - 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{2}{3} sin (\frac{x + 1}{x - 1}))

Take the constant out:

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

= \frac{2}{3} \frac{d}{d x} (sin (\frac{x + 1}{x - 1}))

Apply the chain rule:

cos (\frac{x + 1}{x - 1}) \frac{d}{d x} (\frac{x + 1}{x - 1})

= cos (\frac{x + 1}{x - 1}) \frac{d}{d x} (\frac{x + 1}{x - 1})

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

= \frac{2}{3} cos (\frac{x + 1}{x - 1}) (- \frac{2}{( x - 1 ) ^{2}})

Simplify

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

= - \frac{4 cos ( \frac{x + 1}{x - 1} )}{3 ( x - 1 ) ^{2}}

y^{^{''}} + 4 y^{^{'}} - 21 y = 0

\int csc (t) cot (t) d t

\int sin^{4} (x) d x

arcsin (sin (11 θ))

(\frac{1}{x ^{2}} - \frac{3}{x ^{4}}) (x + 7 x^{3})

What is the derivative of 2/3 sin((x+1/(x-1))) ?
The derivative of 2/3 sin((x+1/(x-1))) is -(4cos(\frac{x+1)/(x-1))}{3(x-1)^2}
What is the first derivative of 2/3 sin((x+1/(x-1))) ?
The first derivative of 2/3 sin((x+1/(x-1))) is -(4cos(\frac{x+1)/(x-1))}{3(x-1)^2}