Question #3c093

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Answer 1 · 2018-01-01T15:58:31

$1/(1+(y/x)^2)*(((dy)/dx*x-1*y)/x^2)$

Use implicit differentiation.
The derivative of $arctan(f(x))$ is $1/(1+(f(x))^2)*f'(x)$ (chain rule)

This means $d/dx(arctan(y/x))=1/(1+(y/x)^2)*d/dx(y/x)$

Using quotient rule:
$=1/(1+(y/x)^2)*(((dy)/dx*x-1*y)/x^2)$

Answer 2 · 2018-01-01T17:17:41

$(d(tan^-1(y/x)))/dx = -y/(x^2+y^2)$

Given: $tan^-1(y/x)$

Using the chain rule, where $u = y/x$

$(d(tan^-1(y/x)))/dx = (d(tan^-1(u)))/(du)(du)/dx$

$(d(tan^-1(y/x)))/dx = 1/(1+u^2)(du)/dx$

$(d(tan^-1(y/x)))/dx = 1/(1+(y/x)^2)(d(y/x))/dx$

$(d(tan^-1(y/x)))/dx = 1/(1+(y/x)^2)(-y/x^2)$

$(d(tan^-1(y/x)))/dx = x^2/(x^2+y^2)(-y/x^2)$

$(d(tan^-1(y/x)))/dx = -y/(x^2+y^2)$