What is the derivative of $y=tan(x+y)$ ?

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Answer 1 · 2016-07-18T13:20:12

$:. dy/dx=-(1+y^2)/y^2$ , or,

$=-csc^2(x+y)$

$y=tan(x+y) rArr tan^-1y=x+y rArr tan^-1y-y=x$

Diff.ing w.r.t. $y$ , we have,

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

$:. (1-1-y^2)/(1+y^2)=dx/dy$

$:.-y^2/(1+y^2)=dx/dy$

$:. dy/dx=1/(dx/dy)=-(1+y^2)/y^2$ , or,

$dy/dx=-{1+tan^2(x+y)}/tan^2(x+y)=-sec^2(x+y)/tan^2(x+y)=-csc^2(x+y)$ #

What is the derivative of y=tan(x+y)y=tan(x+y)?