What is the derivative of (-xsin(x)-cos(x))/(6x^2) ?

The derivative of (-xsin(x)-cos(x))/(6x^2) is -(x^2cos(x)+2(-xsin(x)-cos(x)))/(6x^3)

What is the first derivative of (-xsin(x)-cos(x))/(6x^2) ?

The first derivative of (-xsin(x)-cos(x))/(6x^2) is -(x^2cos(x)+2(-xsin(x)-cos(x)))/(6x^3)

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derivative of

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

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

Solution steps

\frac{d}{d x} (\frac{- x sin ( x ) - cos ( x )}{6 x ^{2}})

Take the constant out:

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

= \frac{1}{6} \frac{d}{d x} (\frac{- x sin ( x ) - cos ( x )}{x ^{2}})

Apply the Quotient Rule:

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

= \frac{1}{6} \cdot \frac{\frac{d}{d x} ( - x sin ( x ) - cos ( x ) ) x ^{2} - \frac{d}{d x} ( x ^{2} ) ( - x sin ( x ) - cos ( x ) )}{( x ^{2} ) ^{2}}

\frac{d}{d x} (- x sin (x) - cos (x)) = - x cos (x)

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

= \frac{1}{6} \cdot \frac{( - x cos ( x ) ) x ^{2} - 2 x ( - x sin ( x ) - cos ( x ) )}{( x ^{2} ) ^{2}}

Simplify

\frac{1}{6} \cdot \frac{( - x cos ( x ) ) x ^{2} - 2 x ( - x sin ( x ) - cos ( x ) )}{( x ^{2} ) ^{2}} : - \frac{x ^{2} cos ( x ) + 2 ( - x sin ( x ) - cos ( x ) )}{6 x ^{3}}

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

\frac{d y}{d x} = x e^{x}

f (x) = - 3 x^{2} - 4 x - 2, at x = 2

\int x x + 1 d x

\int_{0}^{1} 4 cos (\frac{π t}{2}) d t

\int 2 e^{3 x} sin (2 x) d x

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