Question

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`

Answer 1

`(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)`

Answer 2

`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.