QuestionMay 4, 2026

Find the derivative. y=xcoth(2+x^2) y'(x)= square

Find the derivative. y=xcoth(2+x^2) y'(x)= square
Find the derivative. y=xcoth(2+x^2) y'(x)= square

Solution
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Answer

y'(x) = \coth(2 + x^2) - 2x^2 \csch^2(2 + x^2) Explanation 1. Apply product rule Product rule: **(uv)' = u'v + uv'**. Let u = x, v = \coth(2 + x^2). 2. Differentiate u and v u' = 1. v' = -\csch^2(2 + x^2) \cdot (2x) (chain rule used). 3. Combine terms y' = 1 \cdot \coth(2 + x^2) + x \cdot [-\csch^2(2 + x^2) \cdot 2x]. 4. Simplify y' = \coth(2 + x^2) - 2x^2 \csch^2(2 + x^2).

Explanation

1. Apply product rule <br /> Product rule: **$(uv)' = u'v + uv'$**. Let $u = x$, $v = \coth(2 + x^2)$. <br /><br />2. Differentiate $u$ and $v$ <br /> $u' = 1$. <br /> $v' = -\csch^2(2 + x^2) \cdot (2x)$ (chain rule used). <br /><br />3. Combine terms <br /> $y' = 1 \cdot \coth(2 + x^2) + x \cdot [-\csch^2(2 + x^2) \cdot 2x]$. <br /><br />4. Simplify <br /> $y' = \coth(2 + x^2) - 2x^2 \csch^2(2 + x^2)$.
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