$y=x^{-2}+x^3,x \neq0$ 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..
$y=x^{-2}+x^3,x \neq0$ 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..
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
$y \ ' (2) = -2(2)^{-3} + 3(2)^2 = ...$