Find dy/dx : y = sec^-1 (1/2x^2-1), 0< x <1/√2

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asked Jan 18, 2018 in Mathematics by sforrest072 (157,439 points) 61 410 949
edited Mar 7, 2018 by Vikash Kumar

Find dy/dx :

y = sec-1 (1/2x2-1), 0< x <1/√2

1 Answer

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answered Jan 18, 2018 by mdsamim (213,225 points) 5 10 15
selected Jan 18, 2018 by sforrest072
 
Best answer

Differentiating this relationship with respect to x, we obtain

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