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VARVARA [1.3K]
2 years ago
9

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

Mathematics
1 answer:
Vitek1552 [10]2 years ago
5 0

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}}

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