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hw12 derivatives of trigonometric functions (target c3; §3.5) score: 5/7 answered: 5/7 question 6 if ( f(x)=(4x^{2}-2x + 1)cot(x)), find ( f(x)=) question help: message instructor submit question
Step1: Apply product - rule
The product - rule states that if \(y = u\cdot v\), then \(y'=u'v + uv'\). Let \(u = 4x^{2}-2x + 1\) and \(v=\cot(x)\). First, find \(u'\) and \(v'\).
\(u'=\frac{d}{dx}(4x^{2}-2x + 1)=8x-2\)
\(v'=\frac{d}{dx}(\cot(x))=-\csc^{2}(x)\)
Step2: Calculate \(f'(x)\)
Using the product - rule \(f'(x)=u'v+uv'\), we substitute \(u\), \(u'\), \(v\), and \(v'\) into the formula.
\(f'(x)=(8x - 2)\cot(x)+(4x^{2}-2x + 1)(-\csc^{2}(x))=(8x - 2)\cot(x)-(4x^{2}-2x + 1)\csc^{2}(x)\)
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\((8x - 2)\cot(x)-(4x^{2}-2x + 1)\csc^{2}(x)\)