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$y=x^{-2}+x^3,x \neq0$ at the point $(2,8\frac{1}{4})$

i did this way,
derivatative $y'=-2x^{-3}+3x^2$
$8\frac{1}{4}=-2(2)^{-3}+3(2)^2$
$\frac{33}{4}=2(-8)+12$
I don't know how to do further.
the answer is $11\frac{3}{4}$ thanks..

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1 Answer

up vote 1 down vote accepted

The derivative is $y \ '$, not (just) $y$.

So $y \ ' = -2x^{-3} + 3x^2$ tells us that the gradient depends on $x$. Hence we need only use the value $x = 2$ and see what value of $y \ '$ this gives us.

To be more explicit, we could write $y \ '$ as a function of $x$...

$$y \ ' (x) = -2x^{-3} + 3x^2$$

Then

$$y \ ' (2) = -2(2)^{-3} + 3(2)^2 = ...$$

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i got $y'=28$ and....? thx –  Sb Sangpi Apr 13 '12 at 3:45
    
Then you did something wrong! $$y \ '(2) = -2(2)^{-3} + 3(2)^2 = - \dfrac{1}{4} + 12 = 11.75$$ –  The Chaz 2.0 Apr 13 '12 at 3:49
    
thx, my mistake! –  Sb Sangpi Apr 13 '12 at 4:01
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