Question

If log y = tan-¹ x,
Show that (1 + x²) d²y/dx² + (2x-1) dy/dx = 0​

Answer 1

$\\ \red{\bigstar \: \large \underline{ \underline{\bold{Given:- }}}}$

• $\sf \: \log \: y = { \tan}^{ - 1} x$

$\\ \purple{\bigstar \: \large \underline{ \underline{\bold{To \: Show :- }}}}$

• $\sf \: (1 + {x}^{2} ) \dfrac{ {d}^{2}y }{d {x}^{2} } + (2x - 1) \dfrac{dy}{dx} = 0$

$\\ \bigstar \: \large \underline{ \underline{\bold{Solution:- }}}$

$\sf \: \log \: y = { \tan}^{ - 1} x$

Now, Differentiating w.r.t. x, We get

$\\ \sf \: \frac{1}{y} \times \frac{dy}{dx} = \frac{1}{1 + {x}^{2} }$

$\\ \sf \: \: \: \: \implies \dfrac{(1 + {x}^{2} )}{y} \times \frac{dy}{dx} = 0$

$\\ \sf \: \: \: \: \implies {(1 + {x}^{2} )} \times \frac{dy}{dx} = y$

$\\ \sf \: \: \: \: \implies {(1 + {x}^{2} )} \times \frac{ {d}^{2} y}{d {x}^{2} } + 2x \frac{dy}{dx} = \frac{dy}{dx}$

$\\ \sf \: \: \: \: \implies {(1 + {x}^{2} )} \times \frac{ {d}^{2} y}{d {x}^{2} } + (2x - 1) \bigg( \frac{dy}{dx} \bigg)= \frac{dy}{dx}$

$\\ \sf \: \: \: \: \implies {(1 + {x}^{2} )} \times \frac{ {d}^{2} y}{d {x}^{2} } + (2x - 1) \bigg( 2\frac{dy}{dx} - \frac{dy}{dx} \bigg)= 0$

$\\ \sf \: \: \: \: \implies {(1 + {x}^{2} )} \times \frac{ {d}^{2} y}{d {x}^{2} } + (2x - 1) \frac{dy}{dx} - = 0$

Hence proved.

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:logy = {tan}^{ - 1}x$

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

$\rm :\longmapsto\:\dfrac{d}{dx} logy =\dfrac{d}{dx} {tan}^{ - 1}x$

We know that

$\underbrace{ \boxed{ \bf \: \dfrac{d}{dx}logx = \frac{1}{x}}} \: \: \: and \: \: \: \underbrace{ \boxed{ \bf \: \dfrac{d}{dx} {tan}^{ - 1}x = \frac{1}{1 + {x}^{2} }}}$

So, on applying these results, we have

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

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

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

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

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

We know,

$\underbrace{ \boxed{ \bf \: \dfrac{d}{dx}uv = u\dfrac{d}{dx}v + v\dfrac{d}{dx}u}}$

Using this result, we get

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

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

$\rm :\longmapsto\:(1 + {x}^{2})\dfrac{ {d}^{2}y }{d {x}^{2} } + \dfrac{dy}{dx}(2x) - \dfrac{dy}{dx} = 0$

$\rm :\longmapsto\:(1 + {x}^{2})\dfrac{ {d}^{2}y }{d {x}^{2} } + (2x - 1)\dfrac{dy}{dx} = 0$

Hence, Proved

Additional Information :-

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

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

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

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

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

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

$\boxed{ \bf \: \dfrac{d}{dx} {secx} = secx \: tanx}$

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

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