What is the derivative of-1/4 sin(x/2) ?

The derivative of-1/4 sin(x/2) is -1/8 cos(x/2)

What is the first derivative of-1/4 sin(x/2) ?

The first derivative of-1/4 sin(x/2) is -1/8 cos(x/2)

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\frac{d}{d x} (- \frac{1}{4} sin (\frac{x}{2}))

- \frac{1}{8} cos (\frac{x}{2})

Solution steps

\frac{d}{d x} (- \frac{1}{4} sin (\frac{x}{2}))

Take the constant out:

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

= - \frac{1}{4} \frac{d}{d x} (sin (\frac{x}{2}))

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{x}{2}) = \frac{1}{2}

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

Simplify

- \frac{1}{4} cos (\frac{x}{2}) \frac{1}{2} : - \frac{1}{8} cos (\frac{x}{2})

= - \frac{1}{8} cos (\frac{x}{2})

y = 2 x y^{^{'}}

(x + 1) (x^{3} + x - 2)

y^{^{'}} - \frac{6 y}{x} = (\frac{y}{x ^{2}})^{4}

\int \frac{( x + 3 ) ^{4}}{x} d x

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

What is the derivative of-1/4 sin(x/2) ?
The derivative of-1/4 sin(x/2) is -1/8 cos(x/2)
What is the first derivative of-1/4 sin(x/2) ?
The first derivative of-1/4 sin(x/2) is -1/8 cos(x/2)