Q.If $y = \cos(m\cos^{-1}x)$ then show that $(1-x^2)\dfrac{d^2y}{dx^2} - x\dfrac{dy}{dx} + m^2y = 0$

If $y=f(x)$ is differentiable on an interval, its derivative $\dfrac{dy}{dx}=f'(x)$ is itself a function of $x$, and if that function is differentiable too, its derivative is called the second derivative, written $f''(x)$, $y''$, or $\dfrac{d^2y}{dx^2}$. Differentiating again gives the third derivative, and in general the $n$-th derivative $f^{(n)}(x)=\dfrac{d^ny}{dx^n}$ is obtained by differentiating $f(x)$ exactly $n$ times; these are collectively called higher order derivatives. A useful fact: if $f(x)$ is a polynomial of degree $n$, its $n$-th derivative is a nonzero constant and every derivative after that is zero. For functions given parametrically ($x=f(t)$, $y=g(t)$) or implicit …