How do you differentiate $y = {sec}^{- 1} (x^{2} - x)$ $y=sec^-1(x^2-x)$ ?

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Answer 1 · 2018-01-06T09:46:21

$\frac{d y}{d x} = \frac{2 x - 1}{(x^{2} - x) \sqrt{{(x^{2} - x)}^{2} - 1}}$ $(dy)/(dx)=(2x-1)/((x^2-x)sqrt((x^2-x)^2-1)$

$y = {sec}^{- 1} (x^{2} - x)$ $y=sec^(-1)(x^2-x)$

$\Rightarrow x^{2} - x = sec y$ $=>x^2-x=secy$

differentiate $w r t x$ $wrt" "x$

$2 x - 1 = sec y tan y \frac{d y}{d x}$ $2x-1=secytany(dy)/(dx)$

$\frac{d y}{d x} = \frac{2 x - 1}{sec y tan y} - - (1)$ $(dy)/(dx)=(2x-1)/(secytany)--(1)$

now

$1 + {tan}^{2} x = {sec}^{2} x \Rightarrow tan y = \sqrt{({sec}^{2} y - 1)}$ $1+tan^2x=sec^2x=>tany=sqrt((sec^2y-1)$

so substituting back into $(1)$ $(1)$

$\frac{d y}{d x} = \frac{2 x - 1}{(x^{2} - x) \sqrt{{(x^{2} - x)}^{2} - 1}}$ $(dy)/(dx)=(2x-1)/((x^2-x)sqrt((x^2-x)^2-1)$

which can be simplified further as required

How do you differentiate y=sec−1(x2−x)y=sec^-1(x^2-x)?