Question

if r^2 = x^2+y^2 and tan theeta =y/x then prove del theeta/del x = -y/r^2​

Answer 1

$given : \\ {r}^{2} = {x}^{2} + {y}^{2} \\ \tan( \theta) = \frac{y}{x} \\ to \: proof : \\ \frac{d\theta}{dx} = \frac{ - y}{ {r}^{2} }$

$prove : \\ \tan(\theta) = \frac{y}{x} \\ = > y = \tan( \theta ) x \\ square \: both \: side \\ {y}^{2} = \tan {}^{2} (\theta) {x}^{2} \\ so \: \\ {r}^{2} = {x}^{2} + {y}^{2} \\ {r}^{2} = {x}^{2} + \tan {}^{2} (\theta) {x}^{2} \\ {r}^{2} = {x}^{2} (1 + \tan {}^{2} (\theta) ) \\ {r}^{2} = {x}^{2} \sec {}^{2} ( \theta) \\ r = x \sec( \theta)$