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

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Toolbox:
  • $y=f(x)$
  • $\large\frac{dy}{dx}$$=f'(x)$
  • $\large\frac{d^2y}{dx^2}=\frac{d}{dx}\big(\frac{dy}{dx}\big)$
  • $\large\frac{d}{dx}$$(\cos^{-1}x)=\large\frac{-1}{\sqrt{1-x^2}}$
Step 1:
$y=\cos^{-1}x$
Differentiating with respect to $x$
$\large\frac{dy}{dx}=\frac{-1}{\sqrt{1-x^2}}$
$\quad\;=-(1-x^2)^{\large\frac{1}{2}}$
Step 2:
$\large\frac{d^2y}{dx^2}$$=\large\frac{-d}{dx}$$(1-x^2)^{\large\frac{1}{2}}$
$\quad\;=-\big(\large\frac{-1}{2}\big)$$(1-x^2)^{\Large\frac{-3}{2}}$$(-2x)$
$\quad\;=\large\frac{-x}{(1-x^2)^{\Large\frac{3}{2}}}$
$y=\cos^{-1}x$
$x=\cos y$
$\quad\;=\large\frac{-\cos y}{(1-\cos^2y)^{\Large\frac{3}{2}}}$
We know that $\sin^2y+\cos^2y=1$
$\quad\;=\large\frac{-\cos y}{(\sin^2y)^{\Large\frac{3}{2}}}$
answered May 13, 2013 by sreemathi.v
 

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