How do you differentiate $y=cos^-1(1-3x^2)$ ?

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How do you differentiate $y=cos^-1(1-3x^2)$ ?

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Answer 1 · 2016-10-26T19:17:25

$dy/dx=(6x)/sqrt(6x-9x^2)$

Explanation:

Easiest method is to rewrite;

$y=cos^-1(1-3x^2)$ as $cosy=1-3x^2$

and to use implicit differentiation to give:
$-sinydy/dx =-6x$
$:. sinydt/dx =6x$
$:. dy/dx =(6x)/siny$

And using $cos^2A+sin^2A-=1 => siny =sqrt(1-cos^2y)$
$:. siny =sqrt(1-(1-3x^2)^2)$
$:. siny =sqrt(1-(1-3x^2)^2)$
$:. siny =sqrt(1-(1-6x+9x^2))$
$:. siny =sqrt(1-1+6x-9x^2)$
$:. siny =sqrt(6x-9x^2)$

And so, $dy/dx=(6x)/sqrt(6x-9x^2)$

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How do you differentiate $y=cos^-1(1-3x^2)$ ?

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