Question

If log x2 + y2
= tan-1
dy
then show that
dx
y-
ytx​

Answer 1

Correct Statement :-

$\sf \:If \: log( {x}^{2} + {y}^{2}) = 2 {tan}^{ - 1}\dfrac{y}{x} \: then \: show \: \dfrac{dy}{dx} = \dfrac{x + y}{x - y}$

Solution :-

Formula Used :-

$\boxed{ \red{\bf{\dfrac{d}{dx} log(x) = \dfrac{1}{x}}}}$

$\boxed{ \red{\bf{\dfrac{d}{dx} {tan}^{ - 1}x = \dfrac{1}{ {x}^{2} + 1 }}}}$

$\boxed{ \red{\bf{\dfrac{d}{dx} {x}^{2} = 2x }}}$

$\boxed{ \red{\bf{\dfrac{d}{dx}\dfrac{u}{v} = \dfrac{v\dfrac{d}{dx}u - u\dfrac{d}{dx}v}{ {v}^{2} } }}}$

Let's solve the problem now!!

$\rm :\longmapsto\:log( {x}^{2} + {y}^{2}) = 2 {tan}^{ - 1}\dfrac{y}{x}$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}log( {x}^{2} + {y}^{2}) = 2\dfrac{d}{dx} {tan}^{ - 1}\dfrac{y}{x}$

$\rm :\longmapsto\:\: \dfrac{1}{ {x}^{2}+{y}^{2}}\dfrac{d}{dx}( {x}^{2}+{y}^{2}) =2 \dfrac{1}{1 + \dfrac{ {y}^{2} }{ {x}^{2}}}\dfrac{d}{dx}\dfrac{y}{x}$

$\rm\:\dfrac{1}{ \cancel{{x}^{2} + {y}^{2}}}(2x + 2y\dfrac{dy}{dx}) =2 \dfrac{ \cancel {x}^{2} }{ \cancel{{x}^{2}+{y}^{2}}}\bigg(\dfrac{x\dfrac{d}{dx}y - y\dfrac{d}{dx}x}{ \cancel{x}^{2}}\bigg)$

$\rm :\longmapsto\: \cancel2(x + y\dfrac{dy}{dx}) = \cancel2(x\dfrac{dy}{dx} - y)$

$\rm :\longmapsto\:x + y\dfrac{dy}{dx} = x\dfrac{dy}{dx} - y$

$\rm :\longmapsto\:x + y = x\dfrac{dy}{dx} - y\dfrac{dy}{dx}$

$\rm :\longmapsto\:x + y = (x - y)\dfrac{dy}{dx}$

$\bf\implies \:\dfrac{dy}{dx} = \dfrac{x + y}{x - y}$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

$\boxed{ \red{\bf{\dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}}}$

$\boxed{ \red{\bf{\dfrac{d}{dx}k = 0}}}$

$\boxed{ \red{\bf{\dfrac{d}{dx}x = 1}}}$

$\boxed{ \red{\bf{\dfrac{d}{dx}sinx = cosx}}}$

$\boxed{ \red{\bf{\dfrac{d}{dx}cosx = - sinx}}}$

$\boxed{ \red{\bf{\dfrac{d}{dx}tanx = {sec}^{2} x}}}$

$\boxed{ \red{\bf{\dfrac{d}{dx}cotx = { - cosec}^{2} x}}}$

$\boxed{ \red{\bf{\dfrac{d}{dx}secx = secxtanx}}}$

$\boxed{ \red{\bf{\dfrac{d}{dx}cosecx = - cosecx \: cotx}}}$