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 November 3rd, 2010, 09:31 PM #1 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 Questions in derivatives Hi all 1 ) Finf the frisr derivative of y (t^2 + 2/6t)(t^4 - 3 /t^3) my answer : y = t^6 - 3t^2 + 2t^4-6/ 6t^3 =6t^3-6t+t^-3 first derivative 18 - 6 + -3t --------- 2 ) tan(2 + cos3x) derivative = sex^2x =sex2xsin3x 3)sex2x.cos2x derivative = cos2x.-sin2x now appy proudcur rule ?
 November 3rd, 2010, 09:45 PM #2 Senior Member     Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14 Re: Questions in derivatives Oops....translation of your maths language is such an amazing thing!
November 3rd, 2010, 10:31 PM   #3
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Re: Questions in derivatives

Quote:
 Originally Posted by r-soy Hi all 1 ) Finf the frisr derivative of y (t^2 + 2/6t)(t^4 - 3 /t^3) my answer : y = t^6 - 3t^2 + 2t^4-6/ 6t^3 =6t^3-6t+t^-3 first derivative 18 - 6 + -3t --------- 2 ) tan(2 + cos3x) derivative = censored^2x =sex2xsin3x 3)sex2x.cos2x derivative = cos2x.-sin2x now appy proudcur rule ?
1)

Is it $y=(t^2 + \frac{2}{6t})(t^4 - \frac{3}{t^3})$?

first derivative
$y'=(t^2 + \frac{2}{6t})'(t^4 - \frac{3}{t^3})+(t^2 + \frac{2}{6t})(t^4 - \frac{3}{t^3})#39;=(2t - \frac{2}{6t^2})(t^4 - \frac{3}{t^3})+(t^2 + \frac{2}{6t})(4t^3 + \frac{9}{t^4})=6t^5+\frac{3}{t^2}+t^2+\frac{4}{t^5 }$

2)

Is it $\tan{(2 + \cos{3x})}$ ?

$[\tan{(2 + \cos{3x})}]'=\sec^2{(2 + \cos{3x})}(2+\cos{3x})#39;=-3\sec^2{(2 + \cos{3x})}\sin{3x}$

3)

Is it $\sec{2x}\cos{2x}$?

$\sec{2x}\cos{2x}=\frac{1}{\cos{2x}}\cos{2x}=1$,

so the derivative is 0....

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