1. ## Differentiation?

If u = e^(arcsin (y/u)), then prove that d^2y/dx^2 = 2(x^2 + y^2)/(x - y)^3 , where u = √(x^2 + y^2)

2. Hello,
Originally Posted by fardeen_gen
If u = e^(arcsin (y/u)), then prove that d^2y/dx^2 = 2(x^2 + y^2)/(x - y)^3 , where u = √(x^2 + y^2)
This is equivalent to $\ln(u)=\arcsin(y/u)$

-> Implicit differentiation with respect to x...

Left side :
$\frac{\partial}{\partial x} (\ln(u))=\frac{\partial u}{\partial x} \cdot \frac 1u=\left(\frac{\partial}{\partial x} (\sqrt{x^2+y^2})\right) \cdot \frac 1u=\frac{x+yy'}{\sqrt{x^2+y^2}}\cdot \frac 1u=\boxed{\frac{x+yy'}{u^2}}$, where $y'=\frac{dy}{dx}$

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Right side :

$\frac{\partial}{\partial x} (\arcsin(y/u))$ :
By the chain rule, this is :
$\frac{\partial}{\partial x} (y/u) \cdot \frac{1}{\sqrt{1-(y/u)^2}}$

- $\sqrt{1-(y/u)^2}=\sqrt{1-\frac{y^2}{x^2+y^2}}=\sqrt{\frac{x^2}{x^2+y^2}}=\f rac{x}{u}$ (we'll assume x>0...)
- $\frac{\partial}{\partial x} (y/u)=\frac{y'u-y \cdot \frac{\partial u}{\partial x}}{u^2}$
$=\frac{y'u-y \cdot \frac{x+yy'}{u}}{u^2}$

$\Rightarrow \frac{\partial}{\partial x} (\arcsin(y/u))=\frac{y'u-y \cdot \frac{x+yy'}{u}}{u^2} \cdot \frac ux=\frac{y'u^2-y(x+yy')}{xu^2}$

$=\frac{y'x^2+y'y^2-yx-y'y^2}{xu^2}=\frac{y'x^2-yx}{xu^2}=\boxed{\frac{y'x-y}{u^2}}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

So we finally have :

$\frac{x+yy'}{u^2}=\frac{y'x-y}{u^2}$

That is $\boxed{x+yy'=xy'-y}$
(this implies that $y'=\frac{x+y}{x-y}$ - this will be useful for later)

Differentiate implicitely, again with respect to x.
$1+y'^2+yy''=y'+xy''-y'=xy''$, where $y''=\frac{d^2y}{dx^2}$

--> $y''=\frac{1+y'^2}{x-y}$

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

$y''=\frac{(x-y)^2+(x+y)^2}{(x-y)^3}$

$\boxed{y''=\frac{2(x^2+y^2)}{(x-y)^3}}$

(I don't know if there is an easier way to do it though...)