Question

find y. y = \frac{1}{25} \cot (5x + 1) y = \square

Explanation:

Step1: Find the first - derivative \(y'\)

Use the chain rule. The derivative of \(\cot u\) with respect to \(x\) is \(-\csc^{2}u\cdot u'\), where \(u = 5x + 1\) and \(y=\frac{1}{25}\cot(5x + 1)\).
First, \(u'=\frac{d(5x + 1)}{dx}=5\).
Then \(y'=\frac{1}{25}\times(-\csc^{2}(5x + 1))\times5=-\frac{1}{5}\csc^{2}(5x + 1)\).

Step2: Find the second - derivative \(y''\)

Use the chain rule again. The derivative of \(\csc^{2}u\) with respect to \(x\) is \(2\csc u\times(-\csc u\cot u)\cdot u'\).
Here \(u = 5x+1\) and \(u' = 5\).
\(y''=-\frac{1}{5}\times2\csc(5x + 1)\times(-\csc(5x + 1)\cot(5x + 1))\times5\).
Simplify the expression:

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Answer:

\(2\csc^{2}(5x + 1)\cot(5x + 1)\)