What is the derivative of arccos((x+1/3)) ?

The derivative of arccos((x+1/3)) is -1/(sqrt(-x^2-2x+8))

What is the first derivative of arccos((x+1/3)) ?

The first derivative of arccos((x+1/3)) is -1/(sqrt(-x^2-2x+8))

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\frac{d}{d x} (arccos (\frac{x + 1}{3}))

- \frac{1}{- x ^{2} - 2 x + 8}

Solution steps

\frac{d}{d x} (arccos (\frac{x + 1}{3}))

Apply the chain rule:

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

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

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

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

Simplify

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

= - \frac{1}{- x ^{2} - 2 x + 8}

y^{^{'}} = x + y, y (0) = 0

\int u^{2} d u

\int_{0}^{3} 3 x d x

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

\frac{\partial}{\partial y} (x^{\frac{3}{4}} y^{\frac{1}{4}})

What is the derivative of arccos((x+1/3)) ?
The derivative of arccos((x+1/3)) is -1/(sqrt(-x^2-2x+8))
What is the first derivative of arccos((x+1/3)) ?
The first derivative of arccos((x+1/3)) is -1/(sqrt(-x^2-2x+8))