Let f be the function (square root(x^4-16x^2)) find the slope of the line normal to the graph of f at x=6 Please help

Question

Let f be the function (square root(x^4-16x^2)) 

find the slope of the line normal to the graph of f at x=6
Please help ><

rulnick · Answer

To find the slope at x=6, first find the derivative of f (i.e., f'(x)).  Then replace x with 6, and call the resulting value m.
The slope of the line normal to this is then -1/m.

anonymous · Answer

I get how to do it i just don't know if im right because im getting a weird fraction ><

rulnick · Answer

OK, hang on.

anonymous · Answer

alrighty thank you

rulnick · Answer

-sqrt(5)/28, or about -0.080

anonymous · Answer

see i dont get that when i solve

anonymous · Answer

wat did your f'(x) come out to be?

anonymous · Answer

i got 2x^3-16x/(square root ( x^4-16x)

rulnick · Answer

2x(x^2-8)/sqrt(x^4-16x^2)

anonymous · Answer

how do you get that? ><

rulnick · Answer

It looks the same as yours, you just missed the parentheses around the numerator (2x^3-16x).  That may have led to an error on the evaluation at x=6 ... ?

anonymous · Answer

oh okay uhm sec gona check

anonymous · Answer

omg yesh your right i messed up when putting it in thank you so much! ^^

rulnick · Answer

You're very welcome.  Good luck.  Bye.