Question

If 3x2 + y2 = 7 then evaluate the second derivative of y with respect to x when x = 1 and y = 2. Round your answer to 2 decimal places. Use the hyphen symbol, -, for negative values.

Answer 1

start with \[6x+2yy'=0\] then solve for \(y'\)
then repeat

Answer 2

How would I do that? Can you please show me the steps? @satellite73

Answer 3

algebra
subtract \(6x\) from both sides, then divide both sides by \(2y\)

Answer 4

so y= -6x/2y

Answer 5

you might want to reduce

Answer 6

wait is the answer. -6? I replaced x and y with the numbers given?

Answer 7

no that is the first derivative, you need the second

Answer 8

\[y'=-\frac{3x}{y}\] now you need \(y''\)

Answer 9

this one requires the quotient rule

Answer 10

Okay so what is y''?

Answer 11

do you know the quotient rule?

Answer 12

lets put the \(-3\) out front

Answer 13

yes d/dx

Answer 14

\[y'=-3\left(\frac{x}{y}\right)\]\[y''=-3\left(\frac{y-xy'}{y^2}\right)\]

Answer 15

you good from there?

Answer 16

Yes. I plugged in for x and y and I got -2.625 which rounds to -2.63 right?

Answer 17

before you can plug in for x and y you have to replace \(y'\) by \(\frac{-3x}{y}\)

Answer 18

yes I did and I got -1.5

Answer 19

i believe you

Answer 20

Haha so when you plug in y' and x and y I got -2.63. Is that correct?

Answer 21

@satellite73

Answer 22

You there @satellite73

Answer 23

Did you take the second derivative?

Answer 24

I am no sure. I just did what ever satellite73 told me to do. is the second derivative y' ?

Answer 25

y''

Answer 26

the first derivative is \(y'\) the second is \(y''\) but i get the feeling that it was not clear how to find them
true or false?

Answer 27

after you find the first derivative you have to find the second one

Answer 28

true. I found y' which was -1.5 and for y'' I got -2.625.

Answer 29

you want to see how to find them?

Answer 30

yes and is that not correct?

Answer 31

i don't know, i didn't do the arithmetic
but i can show you how to find the first and second derivative if you like

Answer 32

Yes please. Once I find those is y'' my answer?

Answer 33

http://www.wolframalpha.com/input/?i=-3%282-%28-3%29%2F2%29%2F4

Answer 34

looks good

Answer 35

Yay okay so my answer is -2.63 when rounded?

Answer 36

yes

Answer 37

Thank you so much! Can you help me with one more question?

Answer 38

sure if i can

Answer 39

Find the slope of the graph of the relation x2y + 4y = 8 at the point (2, 1).

Answer 40

Differentiate implicitly.

Answer 41

same thing only easier
now lets see how to do it
\(y\) is assumed to be a function of \(x\) so lets suppose it was \(y=\sin(x)\)
what is the derivative of \[x^2\sin(x)+4\sin(x)\]?

Answer 42

x^2cos(x)+2xsin(x)+4cos(x)

Answer 43

exactly

Answer 44

now lets replace \(\sin(x)\) by \(f(x)\) so \(\cos(x)=f'(x)\) what is the derivative of \[x^2f(x)+4f(x)\]?

Answer 45

3x2f+4f

Answer 46

hmm no
think about this answer \[x^2\cos(x)+2x\sin(x)+4\cos(x)
\] and how you used the product rule
replace \)\sin(x)\) by \(f(x)\) and \(\cos(x)\) by \(f'(x)\)

Answer 47

\[x^2\overbrace{\cos(x)}^{f'}+2x\overbrace{\sin(x)}^f+4\overbrace{\cos(x)}^{f'}\]

Answer 48

Okay so x^2f'(x)+2xf(x)+4f'(x)

Answer 49

ok now try i with \(y\) and \(y'\) instead of \(f(x)\) and \(f'(x)\)

Answer 50

\[x^2y'+2xy+4y'\]

Answer 51

right
notice the use of the product rule

Answer 52

and since the derivative of 8 is 0 you have \[x^2y'+2xy+4y'=0\] solve for \(y'\)

Answer 53

I got y'=2x but I don't think that is right.

Answer 54

you there @satellite73

Answer 55

you there @satellite73

Answer 56

oh hold on

Answer 57

no, it is not right

Answer 58

\[x^2y'+2xy+4y'=0\] solve for \(y'\) is like solving \[az+b+cz=0\] for \(z\)

Answer 59

first subtract \(2xy\) get \[x^2y'+4y'=-2xy\] then combine like terms get \[(x^2+4)y'=-2xy\] then divide both sides by \(x^2+4\)

Answer 60

That is where I messed up