If logx2+y2=2tan-1yx, how that dydx=x+yx-y. ORIf xy-yx=ab, find dydx.

Given: logx2+y2=2tan-1yx⇒12logx2+y2=tan-1yxDifferentiate with respect to x, we get,⇒12ddxlogx2+y2=ddxtan-1yx ⇒121x2+y2ddxx2+y2=11+yx2ddxyx ⇒121x2+y2 2x+2ydydx=x2x2+y2 xdydx-yddx(x)x2 ⇒1x2+y2 x+ydydx=x2x2+y2 xdydx-yddx(x)x2 ⇒1x2+y2 x+ydydx=x2x2+y2 xdydx-y(1)x2⇒x+ydydx=xdydx-y ⇒ydydx-xdydx=-y-x ⇒dydx(y-x)=-(y+x) ⇒dydx=-(y+x)y-x ⇒dydx=x+yx-yWhich has proven. OR Given: xy-yx=abLet xy=u and yx=vThen, the function becomes u-v=abdiff w.r.to x on both sidesdudx-dvdx=0 ........(1) u=xy ⇒log u=logxy ⇒log u=y log xNow, differentiate both sides with respect to x, it follows1ududx=log xdydx+y·ddx(log x) ⇒dudx=ulog xdydx+y·1x ⇒dudx=xylog xdydx+yx ...(2)Now, v=yx ⇒log v=logyx ⇒log v=x log yDifferentiate the both side w.r.t. x, it follows,1v·dvdx=log y·ddx(x)+x·ddx(log y) ⇒dvdx=vlog y·1+x·1y·dydx ⇒dvdx=yxlog y+xydydx ....(3)From (1), (2), and (3), we obtainxylog xdydx+yx-yxlog y+xydydx=0 ⇒xylog xdydx-xyx-1dydx+xy-1 y-yx log y=0 ⇒xylog x-xyx-1dydx=yxlog y-xy-1y ⇒dydx=yxlog y-xy-1yxy log x-xyx-1Which is the required answer.

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If $\log (x^{2} + y^{2}) = 2 \tan^{- 1} (\frac{y}{x}),$ how that $\frac{dy}{dx} = \frac{x + y}{x - y} .$
$OR$
If $x^{y} - y^{x} = a^{b},$ find $\frac{dy}{dx} .$

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Detailed Solution

Given: $\log (x^{2} + y^{2}) = 2 \tan^{- 1} (\frac{y}{x})$

$\Rightarrow \frac{1}{2} \log (x^{2} + y^{2}) = \tan^{- 1} (\frac{y}{x})$

Differentiate with respect to x, we get,

$\Rightarrow \frac{1}{2} \frac{d}{dx} \log (x^{2} + y^{2}) = \frac{d}{dx} \tan^{- 1} (\frac{y}{x}) \Rightarrow \frac{1}{2} (\frac{1}{x^{2} + y^{2}}) \frac{d}{dx} (x^{2} + y^{2}) = \frac{1}{1 + {(\frac{y}{x})}^{2}} \frac{d}{dx} (\frac{y}{x}) \Rightarrow \frac{1}{2} (\frac{1}{x^{2} + y^{2}}) [2 x + 2 y \frac{dy}{dx}] = \frac{x^{2}}{(x^{2} + y^{2})} [\frac{x \frac{dy}{dx} - y \frac{d}{dx} (x)}{x^{2}}] \Rightarrow (\frac{1}{x^{2} + y^{2}}) (x + y \frac{dy}{dx}) = \frac{x^{2}}{(x^{2} + y^{2})} [\frac{x \frac{dy}{dx} - y \frac{d}{dx} (x)}{x^{2}}] \Rightarrow (\frac{1}{x^{2} + y^{2}}) (x + y \frac{dy}{dx}) = \frac{x^{2}}{(x^{2} + y^{2})} [\frac{x \frac{dy}{dx} - y (1)}{x^{2}}]$

$\Rightarrow x + y \frac{dy}{dx} = x \frac{dy}{dx} - y \Rightarrow y \frac{dy}{dx} - x \frac{dy}{dx} = - y - x \Rightarrow \frac{dy}{dx} (y - x) = - (y + x) \Rightarrow \frac{dy}{dx} = \frac{- (y + x)}{y - x} \Rightarrow \frac{dy}{dx} = \frac{x + y}{x - y}$

Which has proven.

Given: $x^{y} - y^{x} = a^{b}$

Let $x^{y} = u and y^{x} = v$

Then, the function becomes $u - v = a^{b}$

diff w.r.to x on both sides

$\frac{du}{dx} - \frac{dv}{dx} = 0 . . . . . . . . (1) u = x^{y} \Rightarrow \log u = \log (x^{y}) \Rightarrow \log u = y \log x$

Now, differentiate both sides with respect to x, it follows

$\frac{1}{u} \frac{du}{dx} = \log x \frac{dy}{dx} + y \cdot \frac{d}{dx} (\log x) \Rightarrow \frac{du}{dx} = u [\log x \frac{dy}{dx} + y \cdot \frac{1}{x}] \Rightarrow \frac{du}{dx} = x^{y} (\log x \frac{dy}{dx} + \frac{y}{x}) . . . (2)$

Now,

$v = y^{x} \Rightarrow \log v = \log (y^{x}) \Rightarrow \log v = x \log y$

Differentiate the both side w.r.t. x, it follows,

$\frac{1}{v} \cdot \frac{dv}{dx} = \log y \cdot \frac{d}{dx} (x) + x \cdot \frac{d}{dx} (\log y) \Rightarrow \frac{dv}{dx} = v (\log y \cdot 1 + x \cdot \frac{1}{y} \cdot \frac{dy}{dx}) \Rightarrow \frac{dv}{dx} = y^{x} (\log y + \frac{x}{y} \frac{dy}{dx}) . . . . (3)$

From (1), (2), and (3), we obtain

$x^{y} (\log x \frac{dy}{dx} + \frac{y}{x}) - y^{x} (\log y + \frac{x}{y} \frac{dy}{dx}) = 0 \Rightarrow x^{y} \log x \frac{dy}{dx} - {xy}^{x - 1} \frac{dy}{dx} + x^{y - 1} y - y^{x} \log y = 0 \Rightarrow (x^{y} \log x - {xy}^{x - 1}) \frac{dy}{dx} = y^{x} \log y - x^{y - 1} y \Rightarrow \frac{dy}{dx} = \frac{y^{x} \log y - x^{y - 1} y}{(x^{y} \log x - {xy}^{x - 1})}$