Question

Find the derivatives of the following:

If y = ${\displaystyle \frac{{\sin }^{-1}x}{\sqrt{1-x^2}}}$, show that (1 – x²)y₂ – 3xy₁ – y = 0

Answer 1

y = ${\displaystyle \frac{{\sin }^{-1}x}{\sqrt{1-x^2}}}$

${\displaystyle y\sqrt{1-x^2}={\sin }^{-1}x}$

Differentiating with respect to x

${\displaystyle y\times \frac{1}{2}{(1-x^2)}^{\frac{1}{2}-1}(-2x)+\sqrt{1 x^2}{y}_1=\frac{1}{\sqrt{1-x^2}}}$

${\displaystyle -xy{(1-x^2)}^{-\frac{1}{2}}+\sqrt{1-x^2}{y}_1=\frac{1}{\sqrt{1-x^2}}}$

${\displaystyle {(1-x^2)}^{\frac{1}{2}}[-xy{(1-x^2)}^{-\frac{1}{2}}+{(1-x^2)}^{\frac{1}{2}}{y}_1]}$ = 1

${\displaystyle -xy{(1-x^2)}^{-\frac{1}{2}}\times {(1-x^2)}^{\frac{1}{2}}+{(1-x^2)}^{-\frac{1}{2}} \cdot {(1-x^2)}^{\frac{1}{2}}{y}_1}$ = 1

${\displaystyle -xy+(1-x^2)y_1}$ = 1

Differentiating with respect to x, we get

${\displaystyle -x\cdot y_1+y(-1)+(1-x^2)y_2+y_1(0-2x)}$ = 0

${\displaystyle -x{y}_1-y+(1-x^2)y_2-2x{y}_1}$ = 0

${\displaystyle (1-x^2)y_2-3x{y}_1-y}$ = 0