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How would I find the 2nd derivative of f(X)= (cotx)(cosx)?

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mightyrodrigo | Student | eNotes Newbie

Posted March 4, 2012 at 9:57 AM via web

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How would I find the 2nd derivative of f(X)= (cotx)(cosx)?

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted March 4, 2012 at 10:12 AM (Answer #1)

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The function f(x)= (cot x)(cos x) = `(cos^2 x)/sin x`

f'(x) = `(2*cos x*sin x*sin x - cos^2 x*cos x)/(sin^2x)`

=> `-(2*cos x*sin^2x + cos^3x)/(sin^2 x)`

=> `(-2*cos x - cos^3x)/(sin^2 x)`

f''(x)

=> `2*sin x - (-3*cos^2x*sin x*sin^2x - cos^3 x*2*sin x*cos x)/(sin^4x)`

=> `2 - (-3*cos^2x*sin^2x - cos^3 x*2*cos x)/(sin^3x)`

=> `(2*sin^3x + 3*cos^2x*sin^2x + 2*cos^4 x)/(sin^3x)`

The second derivative of f(x)= (cot x)(cos x)  is f''(x) = `(2*sin^3x + 3*cos^2x*sin^2x + 2*cos^4 x)/(sin^3x)`

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