If y=tan^(-1)(x/y) then dy/dx will be? (a) y/x (b) -y/x (c) x/y (d) -x/y (e) -xy

2 Answers
Jul 27, 2017

(dy)/(dx)=y/(y^2+x^2+x)

Explanation:

As y=tan^(-1)(x/y), we have tany=x/y

and differentiating implicitly

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

or (dy)/(dx)(sec^2y+x/y^2)=1/y

Now as sec^2y=1+tan^2y=1+x^2/y^2

(dy)/(dx)(1+x^2/y^2+x/y^2)=1/y

or (dy)/(dx)((y^2+x^2+x)/y^2)=1/y

or (dy)/(dx)=y/(y^2+x^2+x)

Jul 28, 2017

dy/dx=y/(y^2+x^2+x),

Explanation:

y=tan^-1(x/y) rArr tany=x/y, or, x=ytany.

:. d/dy{x}=d/dy{ytany}.

:. dx/dy=yd/dy(tany)+tany*d/dy(y),........"[Product Rule]"

=ysec^2y+tany=y(1+tan^2y)+tany.

Subst.ing, tany=x/y, in this, we have,

dx/dy=y(1+x^2/y^2)+x/y=y+x^2/y+x/y=(y^2+x^2+x)/y.

rArr dy/dx=y/(y^2+x^2+x), as Respected Shwetank Mauria has

already derived.

No option can be selected for the Answer.