Question

1) Given x^2-16=y^2
a) f''(x) in the simplest factored form
b) f''(5)

Answer 1

f'' is the second derivative so we have to find the first derivative to start

Answer 2

\[x^2 - 16 = y^2\]

Answer 3

That is implicit ^. Would you like to take the derivative of both sides?

Answer 4

yes

Answer 5

Ok, what will it be?

Answer 6

d/dx(x^2-16)=d/dx(y^2)??

Answer 7

Yep, but for the last step ...

Answer 8

\[\frac{ d }{ dx } \space (x^2 - 16) \space = \frac{ d }{ dx } \space (y^2)\]

Answer 9

is it x/y??

Answer 10

Would you like to try the next step? Our first goal is to get dy/dx (that will be f prime)

Answer 11

That's right! It is x/y... And that does make sense, this is a circle

Answer 12

\[f'(x) \space = \frac{ d }{ dx } \space = \frac{ x }{y }\]

Answer 13

so then we need to take the derivative once more to get f'' (second derivative)

Answer 14

how do we do that?

Answer 15

You have two functions x and y that are divided, which means the quotient rule

Answer 16

okay...

Answer 17

\[\frac{ d^2 y }{ dx^2 } \space = \frac{ y(1) - x \frac{ dy }{ dx } }{ y^2 }\]

Answer 18

what do i do after?

Answer 19

ahhh, but you know dy/dx there, so let's substitute that also

Answer 20

\[= \frac{ y - x \space . \frac{ x }{ y } }{ y^2 }\]

Answer 21

and it wants it simplified, alrighty. Any ideas on what do first for that?

Answer 22

get rid of the denominator?

Answer 23

that's what I tried first, too

Answer 24

i'm not sure how to do that

Answer 25

You could get a common denominator in the numerator.
To do that, we'd have to multiply the first fraction by y/y

Answer 26

so first find the common denominator in the numerator

Answer 27

so it would be y^2-x^2/y (1/y^2)

Answer 28

Yes

Answer 29

and then you could multiply that straight across

Answer 30

it would be y^2-x^2/y^3

Answer 31

\[\frac{ y^2 - x^2 }{ y^3 }\]

Answer 32

Yep

Answer 33

Yay! :)

Answer 34

:)

Answer 35

how do i sub in the 5?

Answer 36

Right, because that is only the x = we'd have to solve for the y
using the original equation

Answer 37

So \[5^2 - 16 = y^2\]

Answer 38

so y is plus and minus 3?

Answer 39

yep, I got solutions y = 3, -3

Answer 40

Quick question for the f''(x)
i noticed in the equation of x^2-16=y^2
can be -16=y^2-x^2
so can we say that it's -16/y^3???

Answer 41

Oh, there the left hand side would have a derivative of 0
0 = 2y dy/dx - 2x

Answer 42

because of the constant

Answer 43

so we can't do that?

Answer 44

Right, you can't because you have to take the derivative of both sides separately.
o

Answer 45

Oh, but I see what you mean by plugging in! Hmm, sure... that's fine, actually.

Answer 46

You'll get the same answer either way
\[\frac{ -16 }{ y^3 } \space \]

Answer 47

so for this one, you can't sub in the 5?

Answer 48

oh, that's a good point. There is no x variable left after you simplify, so you would only need to sub y = 3

Answer 49

So it would be 9-x^2/27?

Answer 50

Oh but you know what? y = root (x² - 16) right?
So we could cube that and plug that in then if you wanted, you could sub x = 5 there
using the same ideas you had before like with the
-16 = y² - x2

Answer 51

it would be, yep, but you could keep the -16 in the numerator, too

Answer 52

All of these are correct actually.

Answer 53

Then plugging x = 5 would work, agree

Answer 54

oohhh i see

Answer 55

so for f''(5) answers are plus and minus 16/27??

Answer 56

@Miracrown can you check if it's right? And thank you for helping me out! Also thanks for answering my dummy questions. :)

Answer 57

That looks good to me. :)

Answer 58

:D