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

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Answer 1 · 2017-07-27T07:06:38

$\frac{d y}{d x} = \frac{y}{y^{2} + x^{2} + x}$ $(dy)/(dx)=y/(y^2+x^2+x)$

As $y = {tan}^{- 1} (\frac{x}{y})$ $y=tan^(-1)(x/y)$ , we have $tan y = \frac{x}{y}$ $tany=x/y$

and differentiating implicitly

${sec}^{2} y \frac{d y}{d x} = - \frac{x}{y^{2}} \frac{d y}{d x} + \frac{1}{y}$ $sec^2y(dy)/(dx)=-x/y^2(dy)/(dx)+1/y$

or $\frac{d y}{d x} ({sec}^{2} y + \frac{x}{y^{2}}) = \frac{1}{y}$ $(dy)/(dx)(sec^2y+x/y^2)=1/y$

Now as ${sec}^{2} y = 1 + {tan}^{2} y = 1 + \frac{x^{2}}{y^{2}}$ $sec^2y=1+tan^2y=1+x^2/y^2$

$\frac{d y}{d x} (1 + \frac{x^{2}}{y^{2}} + \frac{x}{y^{2}}) = \frac{1}{y}$ $(dy)/(dx)(1+x^2/y^2+x/y^2)=1/y$

or $\frac{d y}{d x} (\frac{y^{2} + x^{2} + x}{y^{2}}) = \frac{1}{y}$ $(dy)/(dx)((y^2+x^2+x)/y^2)=1/y$

or $\frac{d y}{d x} = \frac{y}{y^{2} + x^{2} + x}$ $(dy)/(dx)=y/(y^2+x^2+x)$

Answer 2 · 2017-07-28T17:12:04

$\frac{d y}{d x} = \frac{y}{y^{2} + x^{2} + x},$ $dy/dx=y/(y^2+x^2+x),$

$y = {tan}^{- 1} (\frac{x}{y}) \Rightarrow tan y = \frac{x}{y}, or, x = y tan y .$ $y=tan^-1(x/y) rArr tany=x/y, or, x=ytany.$

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

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

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

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.

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