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second derivative of 4(x^2-2)^3

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right, i presume u've seen before that you can derive an equation of the form (ax+b)^n, and the answer would be an(ax+b)^(n-1), when you have a higher order polynomial of this form (ax^2+b)^n, you do the chain rule .. u=ax^2+b .. and so on .. there is a quick way to do it in you head which is, write down the bracket, multiply it by the power times by the derivative of what's in the bracket, so in the general case, 2axn(ax^2+b)^(n-1), so for you; y=4(x^2-2)^3 dy/dx= 4*2x*3(x^2-2)^2 dy/dx=24x(x^2-2)^2 you now do it again to get the second derivative, see if you can do it .. hope this helps and it's not too basic and long winded
the first derivative wasnt the hard part. i already did the chain rule and got the first derivative. i just cant get the second
because you have to do the power rule and the chain rule for f"

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Other answers:

no just do the same exact thing again like i've shown you above
96x^2(x^2-2) ??
yh tht's it
sweeet thanks
nws anytime

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