Question

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

Answer 1

Answer:

$y' = -\dfrac{\cot x \csc x}{2 \sqrt{x}}$

Step-by-step explanation:

y = csc x

y' = -cot x csc x

$y = \csc \sqrt{x}$

$y' = \dfrac{d}{dx} [\csc \sqrt{x}]$

$y' = (-\cot x \csc x) \dfrac{d}{dx} \sqrt{x}$

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

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

$y' = -\dfrac{\cot x \csc x}{2 \sqrt{x}}$