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Find the derivative of y with respect to x.
y = csc⁻¹ (x ²+1) , x>0

User Jakozaur
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Final answer:

The derivative of y with respect to x, given y = csc⁻¹ (x²+1) and x > 0, is found by applying the chain rule and the formula for the derivative of the arccosecant function, resulting in dy/dx = -2x / ((x² + 1) ∙ √(x´ + 2x²)).

Step-by-step explanation:

To find the derivative of y with respect to x where y = csc⁻¹ (x²+1), we'll use the chain rule and the formula for the derivative of the arccosecant function.

The derivative of y = csc⁻¹(u) with respect to u is given by dy/du = -1 / (|u| ∙ √(u² - 1)). Applying the chain rule, we get dy/dx = dy/du ∙ du/dx. Here u = x² + 1, so du/dx = 2x.

The resulting derivative is dy/dx = -2x / (|x² + 1| ∙ √((x² + 1)² - 1)). Since x > 0, we can drop the absolute value to get dy/dx = -2x / ((x² + 1) ∙ √(x´ + 2x²)).

User PasNinii
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