Find dy/dx of y=csc(square root of x)

asked Apr 14, 2021 136k views

1 vote

by SiN

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1 vote

Answer:

$y' = -(\cot x \csc x)/(2 √(x))$

Explanation:

y = csc x

y' = -cot x csc x

$y = \csc √(x)$

$y' = (d)/(dx) [\csc √(x)]$

$y' = (-\cot x \csc x) (d)/(dx) √(x)$

$y' = (-\cot x \csc x) (d)/(dx) x^{(1)/(2)}$

$y' = (-\cot x \csc x) (1)/(2) x^{-(1)/(2)}$

$y' = -(\cot x \csc x)/(2 √(x))$

Xxyyxx answered Apr 20, 2021

5.2k points

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