Question

If f(x) = sin¹x, show that(1-x2)f"(x) xf'(x)

Answer 1

$$f(x)=\sin ^{-1}x$$

$$f'(x)=\dfrac{1}{\sqrt{1-x^2}}$$

$$f''(x)=-\dfrac{1}{2(\sqrt{1-x^2})^3}(-2x)=\dfrac{x}{(1-x^2)^{3/2}}$$

$$(1-x^2)f''(x)-xf'(x)$$

$$=(1-x^2)(\dfrac{x}{(1-x^2)^{3/2}})-x(\dfrac{1}{\sqrt{1-x^2}})$$

$$=\dfrac{x}{\sqrt{1-x^2}}-\dfrac{x}{\sqrt{1-x^2}}=0$$