If the solve the problem


If $\log y=\tan ^{-1} x$, show that: $\left(1+x^{2}\right) y_{2}+(2 x-1) y_{1}=0$


Note: $y_{2}$ represents second order derivative i.e. $\frac{d^{2} y}{d x^{2}}$ and $y_{1}=d y / d x$


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

$\therefore y=e^{\tan ^{-1} x} \ldots \ldots$ equation 1

to prove : $\left(1+x^{2}\right) y_{2}+(2 x-1) y_{1}=0$

We notice a second order derivative in the expression to be proved so first take the step to find the second order derivative.

Let's find $\frac{d^{2} y}{d x^{2}}$

As $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$

So, lets first find $d y / d x$

$\frac{d y}{d x}=\frac{d}{d x} e^{\tan ^{-1} x}$

Using chain rule, we will differentiate the above expression

Let $t=\tan ^{-1} x=>\frac{d t}{d x}=\frac{1}{1+x^{2}}\left[\frac{d}{d x} \tan ^{-1} x=\frac{1}{1+x^{2}}\right]$

And $y=e^{t}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dt}} \frac{\mathrm{dt}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{\mathrm{t}} \frac{1}{1+\mathrm{x}^{2}}=\frac{\mathrm{e}^{\tan ^{-1} \mathrm{x}}}{1+\mathrm{x}^{2}} \cdots \cdots$ equation 2

Again differentiating with respect to $x$ applying product rule:

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\mathrm{e}^{\tan ^{-1} \mathrm{x}} \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{1+\mathrm{x}^{2}}\right)+\frac{1}{1+\mathrm{x}^{2}} \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{e}^{\tan ^{-1} \mathrm{x}}$

Using chain rule we will differentiate the above expression-

$\left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}=\frac{e^{\tan ^{-1} x}}{1+x^{2}}-\frac{2 x e^{\tan ^{-1} x}}{1+x^{2}}$

$\left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}=\frac{e^{\tan ^{-1} x}}{1+x^{2}}(1-2 x)$

Using equation 2 :

$\left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}=\frac{d y}{d x}(1-2 x)$

$\therefore\left(1+x^{2}\right) y_{2}+(2 x-1) y_{1}=0 \ldots \ldots$ proved

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