What is the derivative of arccos(y/x) ?

The derivative of arccos(y/x) is -(x\frac{dy)/(dx)-y}{xsqrt(x^2-y^2)}

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

The first derivative of arccos(y/x) is -(x\frac{dy)/(dx)-y}{xsqrt(x^2-y^2)}

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\frac{d}{d x} (arccos (\frac{y}{x}))

- \frac{x \frac{d y}{d x} - y}{x x ^{2} - y ^{2}}

Solution steps

\frac{d}{d x} (arccos (\frac{y}{x}))

Apply the chain rule:

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

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

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

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

Simplify

- \frac{1}{1 - ( \frac{y}{x} ) ^{2}} \cdot \frac{x \frac{d y}{d x} - y}{x ^{2}} : - \frac{x \frac{d y}{d x} - y}{x x ^{2} - y ^{2}}

= - \frac{x \frac{d y}{d x} - y}{x x ^{2} - y ^{2}}

y^{^{'}} = y \cdot tan (x) + sin (x)

\frac{\partial}{\partial x} (x ln (x^{2} y + y^{2}))

\int 3 x^{2} (x^{3} + 1)^{6} d x

\int x (x + 1)^{- 4} d x

\int (x^{8} + sec^{2} (x)) d x

What is the derivative of arccos(y/x) ?
The derivative of arccos(y/x) is -(x\frac{dy)/(dx)-y}{xsqrt(x^2-y^2)}
What is the first derivative of arccos(y/x) ?
The first derivative of arccos(y/x) is -(x\frac{dy)/(dx)-y}{xsqrt(x^2-y^2)}