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)#