Question

Please help: let f be the function given by
a) Find the domain of f.
b) Find f '(x)
c) Find the slope of the line normal to the graph of f at x = 5

f(x)=((x^4)- (16x^2))^(1/2)

Answer 1

chain rule; other than that its pretty basic

Answer 2

say u = that middle stuff; then

\[D\sqrt{u} = \frac{u'}{2\sqrt{u}}\]

Answer 3

the original equation is all under a radical sign.. do i still set it up how i posted it?

Answer 4

yeah, ^1/2 is fine for radicals

Answer 5

some even perfer it when doing derivatives, helps them keep track of the "power" rule

Answer 6

why is the denominator 2 radical u?

Answer 7

becasue I cleaned it up by saying: u = that stuff in the middle ....

Answer 8

\[D[u^{1/2}]=\frac{1}{2}u'u^{-1/2}\]

Answer 9

oh . now i got it!

Answer 10

fill in the x parts and solve it for x=5

Answer 11

Find the slope of the line [normal] to the graph

they try to trick you here

Answer 12

the slope is y=mx+b ?

Answer 13

I have another question: that chain rule, i tought is was au^a-1

Answer 14

it is

Answer 15

the slope IS f'(x)

the normal is the slope perp to that: or simply, -1/f'(x)

Answer 16

ok for the chain rule D[u^1/2]= 1/2 u^(-1/2)

Answer 17

yes

Answer 18

well, since we are using u as a function of x; u(x) we pop out a derivative as well

Answer 19

now i think i might be lost

Answer 20

wait do you mean derivative of everything under the radical

Answer 21

does u represent a function of x?

Answer 22

u = x^4 - 16x^2

right?

Answer 23

yeah im pretty positive
thats under the radical

Answer 24

then since u is a function of x and not just a variable in and of itself; anything done to u pops out its derivative

the derivative rules SHOULD be taught as functions in their own right so that you dont have to "Add to" what youve already been taught with extra memory :)

Answer 25

ok that makes much sense

Answer 26

\[\frac{d}{dx}[u(x)]^n=nu(x)^{n-1}*u'(x)\]

Answer 27

the chain rule is always working within these other rules .... always

youre just not taught that since x' = 1 when it pops out

Answer 28

\[D[x^2]=2x*x'\]
but since x'=1 it seems rather redundant to do it that way

Answer 29

ok i totally understand. now my answer will have no variables right?

Answer 30

that is if i plug 5 into x

Answer 31

itll have an x in there to plug in the 5; but yes, the slope itself and the normal produced will have no variables for a given point