What is the derivative of ${tan}^{- 1} (\frac{y}{x}) = 0$ $Tan^-1 (y/x) = 0$ ?

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Answer 1 · 2017-08-03T13:13:13

$\frac{d y}{d x} = 0$ $dy/dx = 0$ provided $x \neq 0$ $x != 0$

${tan}^{- 1} (\frac{y}{x})$ $tan^(-1)(y/x)$ is only defined if $x \neq 0$ $x != 0$

In which case we have:

${tan}^{- 1} (\frac{y}{x}) = 0$ $tan^(-1)(y/x) = 0$
$:. y/x = 0$
$:. y=0 => dy/dx = 0$

What is the derivative of Tan^-1 (y/x) = 0tan−1(yx)=0 Tan^-1 (y/x) = 0?