If y = cos^{-1} x,find frac{d^{2} y}{d x^{2}} | vedclass

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Question

(A) Given $y = \cos^{-1} x$,which implies $x = \cos y$.Differentiating with respect to $x$:$\frac{dy}{dx} = -\frac{1}{\sqrt{1-x^2}} = -(1-x^2)^{-1/2}$

Vedclass · Accepted Answer

$-\cot y \cdot \csc^{2} y$

Vedclass · Answer

(A) Given $y = \cos^{-1} x$,which implies $x = \cos y$.Differentiating with respect to $x$:$\frac{dy}{dx} = -\frac{1}{\sqrt{1-x^2}} = -(1-x^2)^{-1/2}$.Now,differentiating again with respect to $x$:$\frac{d^2y}{dx^2} = \frac{d}{dx} [-(1-x^2)^{-1/2}]$$= -(-\frac{1}{2})(1-x^2)^{-3/2} \cdot (-2x)$$= -\frac{x}{(1-x^2)^{3/2}}$.Since $x = \cos y$,then $1-x^2 = 1-\cos^2 y = \sin^2 y$.Substituting these into the expression:$\frac{d^2y}{dx^2} = -\frac{\cos y}{(\sin^2 y)^{3/2}}$$= -\frac{\cos y}{\sin^3 y}$$= -\frac{\cos y}{\sin y} \cdot \frac{1}{\sin^2 y}$$= -\cot y \cdot \csc^2 y$.

If $y = \cos^{-1} x$,find $\frac{d^{2} y}{d x^{2}}$ in terms of $y$ alone.

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