What is the derivative of 3arccos(x/8) ?

The derivative of 3arccos(x/8) is -3/(sqrt(64-x^2))

What is the first derivative of 3arccos(x/8) ?

The first derivative of 3arccos(x/8) is -3/(sqrt(64-x^2))

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\frac{d}{d x} (3 arccos (\frac{x}{8}))

- \frac{3}{64 - x ^{2}}

Solution steps

\frac{d}{d x} (3 arccos (\frac{x}{8}))

Take the constant out:

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

= 3 \frac{d}{d x} (arccos (\frac{x}{8}))

Apply the chain rule:

- \frac{1}{1 - ( \frac{x}{8} ) ^{2}} \frac{d}{d x} (\frac{x}{8})

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

\frac{d}{d x} (\frac{x}{8}) = \frac{1}{8}

= 3 - \frac{1}{1 - ( \frac{x}{8} ) ^{2}} \cdot \frac{1}{8}

Simplify

3 - \frac{1}{1 - ( \frac{x}{8} ) ^{2}} \cdot \frac{1}{8} : - \frac{3}{64 - x ^{2}}

= - \frac{3}{64 - x ^{2}}

\frac{d y}{d x} = 3 + y - 3 x + 2

y = \frac{1}{x + 1}, y = x + 2, x = 2

\frac{d}{d x} ((x + 1)^{x})

\int \frac{( y ^{4} + 2 y ^{2} - 1 )}{y} d y

\int - \frac{1}{1 + e ^{x}} d x

What is the derivative of 3arccos(x/8) ?
The derivative of 3arccos(x/8) is -3/(sqrt(64-x^2))
What is the first derivative of 3arccos(x/8) ?
The first derivative of 3arccos(x/8) is -3/(sqrt(64-x^2))