evaluate y^ii(2),when y=x^-2 a 0.375 b 0.148 c 0.268 d 0.425
If y = 1/x^2, what is y' and y'' in general? That's the first step.
0/0 but i see y^ii (2) how about that?
Still there? 0/0 has nothing to do with it. The first derivative is \[ y' = -2.x^{-3} \] What's the second derivative?
6x
No, \[ y'' = (-3)(-2)x^{-4} = 6x^{-4} \] Hence you can now evaluate \[ y''(2) \] by substituting \( x = 2 \) into the formula for the second derivative.
wow thanks the answer is near letter d is this correct?
No. Substitute x=2 into that equation and you have \[ y''(2) = \frac{6}{16} = \frac{3}{8} \] Which decimal answer is that equal to?
Are you following?
0.375
sorry i thought what i did is my prof do.he times 6 in the sqrt because for him x=1
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