If logx2-y2-2 tan -1y-x- show that d y-d x-x-y-x-y-

Log (x^2+y^2)=2tan^ 1(yx) Differentiating with respect to x, we get 1x^2+y^2(2x+2ydydx)=2×1/1+(yx)^2×xdydx y× 1x^2 ⇒1x^2+y^2(2x+2ydydx)=2×x^2x^2+y^2×xdydx yx^2 ...

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Byju's Answer

Standard XII

Mathematics

Global Maxima

If logx2+y2=2...

Question

$If log (x^{2} + y^{2}) = 2 {tan}^{- 1} (\frac{y}{x}),$ show that $\frac{d y}{d x} = \frac{x + y}{x - y} .$

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Solution

$log (x^{2} + y^{2}) = 2 {tan}^{- 1} (\frac{y}{x})$
Differentiating with respect to $x$ , we get
$\frac{1}{x^{2} + y^{2}} (2 x + 2 y \frac{d y}{d x}) = 2 \times \frac{1}{1 + {(\frac{y}{x})}^{2}} \times \frac{x \frac{d y}{d x} - y \times 1}{x^{2}}$
$\Rightarrow \frac{1}{x^{2} + y^{2}} (2 x + 2 y \frac{d y}{d x}) = 2 \times \frac{x^{2}}{x^{2} + y^{2}} \times \frac{x \frac{d y}{d x} - y}{x^{2}}$
$\Rightarrow 2 x + 2 y \frac{d y}{d x} = 2 (x \frac{d y}{d x} - y)$
$\Rightarrow x + y \frac{d y}{d x} = (x \frac{d y}{d x} - y)$
$\Rightarrow \frac{d y}{d x} (x - y) = x + y$
$\Rightarrow \frac{d y}{d x} = \frac{x + y}{x - y}$

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If log(x2+y2)=2tan−1(yx), show that dydx=x+yx−y.

log(x2+y2)=2tan−1(yx) Differentiating with respect to x, we get 1x2+y2(2x+2ydydx)=2×11+(yx)2×xdydx−y×1x2 ⇒1x2+y2(2x+2ydydx)=2×x2x2+y2×xdydx−yx2 ⇒2x+2ydydx=2(xdydx−y) ⇒x+ydydx=(xdydx−y) ⇒dydx(x−y)=x+y ⇒dydx=x+yx−y

$If log (x^{2} + y^{2}) = 2 {tan}^{- 1} (\frac{y}{x}),$ show that $\frac{d y}{d x} = \frac{x + y}{x - y} .$