Question

Help! i need to find all the second partial derivatives

Answer 1

z= y/(2x+3y)

Answer 2

i am not sure how to do this one

Answer 3

Math isn't my thing, but I'm taking some math classes right now. I think this is right.

Answer 4

@jim_thompson5910

Answer 5

yeah but how did you get that?

Answer 6

Honestly I don't even know if it is right. Let me make sure its right before I try to explain

Answer 7

have you tried using quotient rule yet?

Answer 8

\[\text{ Let } f=y \text{ and } g=2x+3y \\ \text{ do you know how to find the following: } \\ f_x , g_x , f_y,g_y\]
if so can you please state them each

Answer 9

fx=0 fy=1 gx=2 gy=3 i think

Answer 10

that's right
\[f=f(x,y) \text{ and } g=g(x,y) \\ z=\frac{f}{g} \\ z_x=\frac{f_x \cdot g-g_x \cdot f}{g^2} \\ z_y=\frac{f_y \cdot g-g_y \cdot f}{g^2}\]
can you find these?
this is by quotient rule

Answer 11

that is z_x and z_y

Answer 12

okay let me work it out

Answer 13

okay for zx i got 2y/(2x+3y)^2 and for zy i got 2x/(2x+3y)^2

Answer 14

\[z_x=\frac{0 \cdot (2x+3y)-2 \cdot y}{(2x+3y)^2} =\frac{-2y}{(2x+3y)^2} \\ z_y=\frac{1 \cdot (2x+3y)-3 \cdot y}{(2x+3y)^2}=\frac{2x+3y-3y}{(2x+3y)^2}=\frac{2x}{(2x+3y)^2}\]

Answer 15

now for second partial derivatives do i do the same thing?

Answer 16

you apply quotient rule yes
\[(\frac{u}{v})_k=\frac{u_k \cdot v-v_k \cdot u}{v^2}\]

Answer 17

before you do that let me ask you if you know how to find the following:
\[\text{ Let } u=-2y \text{ and } v=(2x+3y)^2 \\ \text{do you know how \to find the following } \\ u_x , u_y, v_x, v_y\]

Answer 18

let see ux= 0 uy=-2 vx=4(2x+3y) vy=6(2x+3y)

Answer 19

are they right?

Answer 20

omg you did it
I was kinda wondering if you would do the chain rule correctly but you did
great job

Answer 21

yes!

Answer 22

there is one other problem im stuck on if you could help me with real quick its v=sin(s^2-t^2)

Answer 23

\[z_{xx}=(z_x)_x=\frac{u_x \cdot v-v_x \cdot u}{v^2} \\ z_{xy}=(z_x)_y=\frac{u_y \cdot v-v_y \cdot u}{v^2}\]
and okay ... I guess you are able to finish this other problem now?

Answer 24

i know that sin changes to cos but the inside part is confusing when looking for the second partial derivatives

Answer 25

\[v=\sin(s^2-t^2) \\ v_s=(s^2-t^2)_s \cos(s^2-t^2) \\ v_t=(s^2-t^2)_t \cos(s^2-t^2)\]
so you have gotten this far?

Answer 26

(s^2-t^2)_s and the (s^2-t^2)_t still need to be evaluated

Answer 27

yes

Answer 28

so you know
\[v_s=2s \cos(s^2-t^2) \\ \text{ and } v_t=-2t \cos(s^2-t^2)\]

Answer 29

to find v_ss you will need product rule

Answer 30

v_st you do not need product rule

Answer 31

\[v_{ss}=(2s)_s \cos(s^2-t^2)+2s[ \cos(s^2-t^2)]_s\]

Answer 32

yeah i got vs but why dont you need the product rule for v_st?

Answer 33

because you have a function s times a function of t

Answer 34

you can treat the function of s as a constant multiple since it is to be treated independently of t in partials

Answer 35

when it comes to finding that partial w.r.t. t of course

Answer 36

\[v_{st}=2s[\cos(s^2-t^2)]_t\]
you could use product rule but you don't need it
if you used product rule instead of constant multiple rule it would look like this:
\[v_{st}=[2s]_t \cos(s^2-t^2)+2s[\cos(s^2-t^2)]_t \\ \text{ but notice this is the same thing I just wrote since } [2s]_t=0\]

Answer 37

oh okay

Answer 38

so would v_st and v_ts be the same?

Answer 39

I was trying to prove it but it was taking too long
but yeah they are
http://tutorial.math.lamar.edu/Classes/CalcIII/HighOrderPartialDerivs.aspx

Answer 40

i still dont get it so you only have to derive functions of t when its ts and derive functions of s when its st?

Answer 41

not sure what you mean

Answer 42

\[z_{ts}=(z_t)_s \\ \text{ means we are differentiating } z_t \text{ with respect to } s\]

Answer 43

\[z_{st}=(z_{s})_t \\ \text{ means we are differentiating } z_s \text{ with respect to } t \]

Answer 44

could you show that with the problem we were working on?

Answer 45

so did you understand what I wrote eariler about z_st and z_ts ?

Answer 46

sort of

Answer 47

\[v_s=2s \cos(s^2-t^2) \\ \text{ and } v_t=-2t \cos(s^2-t^2) \]

\[v_{st}=2s[\cos(s^2-t^2)]_t \\ v_{ts}=-2t[\cos(s^2-t^2)]_s\]

Answer 48

I used constant multiple rule

Answer 49

then you just need chain rule

Answer 50

okay so for v_st you get (4st)sin(s^2-t^2) cause you're only differentiating t right?

Answer 51

right you differentiating the function called v_s w.r.t. t

Answer 52

and same thing for v_ts = 4s^2sin(s^2-t^2) cause you're only differentiating s right?

Answer 53

oh no where you get s^2?

Answer 54

s^2 = 2s

Answer 55

yes but why is there an extra s

Answer 56

\[v_t=-2t \cos(s^2-t^2) \\ v_{ts}=(v_t)_s=(-2t \cos(s^2-t^2))_s \\ \\ v_{ts}=-2t \cdot (2s) \cdot -\sin(s^2-t^2) \\ \\ v_{ts}=4 ts \sin(s^2-t^2)\]

Answer 57

oh sorry i made a mistake

Answer 58

so yeah its 4ts sin(s^2-t^2)

Answer 59

so we use the first partial derivatives to find v_st and v_ts

Answer 60

right?

Answer 61

we use the first partials always to find the second partials

Answer 62

even if we were finding v_ss or v_tt

Answer 63

we still have to use v_s and v_t to find those

Answer 64

okay i think understand it a little bit more

Answer 65

example

\[v=st^2 +s+t^2\\ v_s=t^2+1 \\ v_t =2s t+2t \\ \text{ to find } v_{st} \text{ we use } v_s \\ v_{st}=(v_s)_t=(t^2+1)_t=2t \\ \text{ to find } v_{ss} \text{ we use } v_s \\ v_{ss}=(v_s)_s=(t^2+1)_s=0 \\ \text{ to find } v_{ts} \text{ we use } v_t \\ v_{ts}=(v_t)_s=(2st+2t)_s=2t \\ \text{ to find } v_{tt} \text{ we use} v_t \\ v_{\tt}=(v_t)_t=(2st+2t)_t=2s+2\]

Answer 66

oh i see it now

Answer 67

i know what to do now thank you @freckles your a lifesaver

Answer 68

np

Answer 69

I made a type-o

Answer 70

left off the subscript (tt) on that last line on the first part

Answer 71

\[v=st^2 +s+t^2\\ v_s=t^2+1 \\ v_t =2s t+2t \\ \text{ to find } v_{st} \text{ we use } v_s \\ v_{st}=(v_s)_t=(t^2+1)_t=2t \\ \text{ to find } v_{ss} \text{ we use } v_s \\ v_{ss}=(v_s)_s=(t^2+1)_s=0 \\ \text{ to find } v_{ts} \text{ we use } v_t \\ v_{ts}=(v_t)_s=(2st+2t)_s=2t \\ \text{ to find } v_{\tt} \text{ we use} v_t \\ v_{tt}=(v_t)_t=(2st+2t)_t=2s+2\]