Secants to tangents: dy/dx

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

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

Default solution

Given $y = \sin^{-1} x$, show that $(1-x^2)\displaystyle\frac{\mathrm{d}^3y}{dx^3} - 3x\frac{\mathrm{d}^2y}{dx^2} - \frac{\mathrm{d}y}{\mathrm{d}x} = 0$

Diff to $x$:
$\displaystyle\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\sqrt{1-x^2}}$

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

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

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

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

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

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

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

Practice 4 Q20

Practice 4 Q20

Video Solution:

Default solution