Proving Second Derivative Of Arcsin(x)
Given that $y={{\sin }^{-1}}x$, prove that $\left( 1-{{x}^{2}} \right)\frac{{{\mathrm{d}}^{3}}y}{\mathrm{d}{{x}^{3}}}-3x\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}-\frac{\mathrm{d}y}{\mathrm{d}x}=0$.

**Proving Second Derivative Of Arcsin(x)**

Given that $y={{\sin }^{-1}}x$, prove that $\left( 1-{{x}^{2}} \right)\frac{{{\mathrm{d}}^{3}}y}{\mathrm{d}{{x}^{3}}}-3x\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}-\frac{\mathrm{d}y}{\mathrm{d}x}=0$.

Techniques of Differentiation - General

Techniques of Differentiation

H2 Math

International Baccalaureate Higher Level Mathematics

IB HL Math

Proving Third Order Differential Equation of arcsin (x)

$y = \sin^{-1}x$

Differentiate with respect to $x$:
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\sqrt{1-x^2}}$

Square both sides:
$\left(\sqrt{1-x^2}\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = 1$

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

Differentiate w.r.t. $x$ using product rule:
$(1-x^2) \cdot 2\frac{\mathrm{d}y}{\mathrm{d}x}\frac{\mathrm{d}^2y}{dx^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2(-2x) = 0$

Divide by $2\frac{\mathrm{d}y}{\mathrm{d}x}$:
$(1-x^2)\frac{\mathrm{d}^2y}{dx^2} - x\frac{\mathrm{d}y}{\mathrm{d}x} = 0$

Differentiate w.r.t. $x$ again:
$(1-x^2)\frac{\mathrm{d}^3y}{dx^3} + \frac{\mathrm{d}^2y}{dx^2}(-2x) - \left[x\frac{\mathrm{d}^2y}{dx^2} + \frac{\mathrm{d}y}{\mathrm{d}x}\right] = 0$

$(1-x^2)\frac{\mathrm{d}^3y}{dx^3} - 3x\frac{\mathrm{d}^2y}{dx^2} - \frac{\mathrm{d}y}{\mathrm{d}x} = 0$ (as shown)

Proving Third Order Differential Equation of arcsin (x)

Proving Third Order Differential Equation of arcsin (x)

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