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# `1.f(x)=(x^3-6)/(x^2)` Find the Derivative of Function.``2.f(x)=`root(3)(x)+root(5)(x)`

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`1.f(x)=(x^3-6)/(x^2)`

Find the Derivative of Function.

``2.f(x)=`root(3)(x)+root(5)(x)`

Posted by stevenbawi103 on February 14, 2013 at 8:14 PM via web and tagged with caculus 1, math

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`f(x)= (x^3-6)/x^2`

To take the derivative of the function, apply the quotient rule `(u/v)'=(v*u'-u*v')/v^2` .

So, let  `u = x^3 - 6`    and   `v=x^2` .

Then, take the derivative of u and v.

`u'=(x^3-6)'=(x^3)'-6' = 3x^2-0=3x^2`

`v'=(x^2)'=2x`

And, plug-in u, v, u' and v' to the formula.

`f'(x)= ((x^3-6)/x^2)' = (x^2*3x^2-(x^3-6)*2x)/(x^2)^2`

Next, simplify f'(x).

`f'(x)=(3x^4-(2x^4-12x))/x^4=(3x^4-2x^4+12x)/x^4=(x^4+12x)/x^4`

`f'(x)=(x(x^3+12))/x^4=(x^3+12)/x^3`

Hence, the derivative of  `f(x)= (x^3-6)/x^2`  is  `f'(x)=(x^3+12)/x^3` .

Posted by mjripalda on February 15, 2013 at 2:45 AM (Answer #1)