Question

Implicit differentiation question..
Topic: Related Rates
Suppose that x(t) and y(t) are related by xy = y^4-x for all *t* and t = 1, x(1) =0.5, y(1) = 1 and dx/dt = 7. Find dy/dt.

Answer 1

https://imgur.com/3sp2d4D

Answer 2

I believe the implicit differentiation is :
xy = y^4 - x
y dx/dt + x dy/dt = 4y^3 dydt - dy/dx

Answer 3

try to simplify first then do substitution

Answer 4

https://prnt.sc/phg571

Answer 5

\[\rm x\cdot y=y^4-x\]
\[\rm x\frac{dy}{dt}+y\frac{dx}{dt}=4y^3\frac{dy}{dt}-1\frac{dx}{dt}\]

Answer 6

\[\rm x\cdot y=y^4-x\]\[\rm x\color{red}{\frac{dy}{dt}}+y\color{blue}{\frac{dx}{dt}}=4y^3\color{Red}{\frac{dy}{dt}}-1\color{blue}{\frac{dx}{dt}}\]
like terms on same side
\[\rm y\color{blue}{\frac{dx}{dt}}+1\color{blue}{\frac{dx}{dt}}=4y^3\color{red}{\frac{dy}{dt}}-x\color{Red}{\frac{dy}{dt}}\]\[\rm (y+1)\color{blue}{\frac{dx}{dt}}=(4y^3-x)\color{Red}{\frac{dy}{dt}}\]
\[\rm\color{reD}{\frac{ dy }{ dt }}=(\frac{y+1}{4y^3-x})\color{blue}{\frac{dx}{dt}}\]

Answer 7

When you take the derivative of *y*, don’t you write dy/dx??? In this case it’d be dy/dt.... when did you write dx/dt when took the derivative of *y* and same with *x*.... you wrote dy/dt when you took the derivative of x....shouldn’t it be dx/dt??? I’m conpuzzeled :/

Answer 8

there are so many x's and dx's/dt which one?

Answer 9

Let me write....

Answer 10

apply the product rule??

Answer 11

Yes i know how to apply the product rule :) but i’m Not familiar with what you did with switching the dy/dt when taking the derivative of x and vice versa with y.....dx/dt....

Answer 12

When i take the derivative of x...i write dx/dt with it and same with y...i write dy/dt with it

Answer 13

\[\rm \color{Red}{f(x)}\cdot\color{blue}{ g(x)}= f(x)'\color{blue}{g(x)}+\color{red}{f(x)}g(x)'\]\[\rm \color{Red}{x}\cdot\color{blue}{ y}= x'\color{blue}{y}+\color{red}{x}y'\]
so derivative of x--> 1dx/dt
derivative of y-->1 dy/dt
\[\rm \color{Red}{x}\cdot\color{blue}{ y}= \frac{dx}{dt}\cdot \color{blue}{y}+\color{red}{x}\cdot \frac{dy}{dt}\]

Answer 14

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Ballery1
When i take the derivative of x...i write dx/dt with it and same with y...i write dy/dt with it
\(\color{#0cbb34}{\text{End of Quote}}\)
yes right. but you're not taking the derivative of `x` and `y` which is just 1.

Answer 15

for example
\[\rm (y^4 )'\]
you don't just write \[\rm y^4 \frac{dy}{dt}\]
you have to take the derviative of `y^4` which is 4y^3

Answer 16

Yes...

Answer 17

Ok do you want me to write out how i’d Take the derivative of the whole equation??? You can just correct me where i’m Wrong.??? Sound coo??

Answer 18

okay.

Answer 19

Uhggg....u caught me

Answer 20

same as what I did.

Answer 21

K just tell me why do you write dy/dt when you take the derivative of x :/ i’m Not getting it...

Answer 22

where??

Answer 23

this one?

Answer 24

:/

Answer 25

i just showed you.

Answer 26

I don’t understand.... aww ;-; ... i wanna cry :/

Answer 27

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Nnesha
\[\rm \color{Red}{f(x)}\cdot\color{blue}{ g(x)}= f(x)'\color{blue}{g(x)}+\color{red}{f(x)}g(x)'\]\[\rm \color{Red}{x}\cdot\color{blue}{ y}= x'\color{blue}{y}+\color{red}{x}y'\]
so derivative of x--> 1dx/dt
derivative of y-->1 dy/dt
\[\rm \color{Red}{x}\cdot\color{blue}{ y}= \frac{dx}{dt}\cdot \color{blue}{y}+\color{red}{x}\cdot \frac{dy}{dt}\]
\(\color{#0cbb34}{\text{End of Quote}}\)
here why do i have have `dy/dt` with x is because i'm applying the product rule
I am NOT taking the derviative of x that's why it still there

Answer 28

So when you take the derivative of one variable ..let’s say x’...you replace it with dx/dy???

Answer 29

Give me a sample question to work on..

Answer 30

\[x'= 1 \cdot \ \frac{dx}{dx} = 1\]

Answer 31

Yes that’s true...

Answer 32

I see now wat iz confusing u...

Answer 33

Assuming x and y = 1

Answer 34

From there i plugin the values of y, x and dy/dx.... and solve for the required variable....

Answer 35

yes but what about `t` ??

Answer 36

taking derivative with respect to `t`

Answer 37

Replace x with t

Answer 38

so dx/dx-->dx/dt

Answer 39

Yes...

Answer 40

\[\rm 1 \frac{dx}{dt}\cdot y+x \cdot 1 \frac{dy}{dt}\]

Answer 41

Yes yes yes .... dx/dt,,,

Answer 42

Actually i think i got it... i’ll Git back to you if i have trouble finishing up the question... i’m Post my solution when i’m Done... thank you shooooooooooooo much :)

Answer 43

No flirts intended..

Answer 44

good luck.

Answer 45

Screw the f(x).g(x) + f(x). G’(x) method.. i’m Plugging in the variables straight into the product rule..

Answer 46

And from there.. i’m Just gonna label what’s dx/dt and what’s dy/dt

Answer 47

you're on the right track. you know how to take derivatives just keep in mind for this question you're taking derviative with respect to t not x

Answer 48

do you have the key for this q?

Answer 49

I’m still getting freakin 3...instead of 4.... ugh.... /.-

Answer 50

The answer is 4... -_-

Answer 51

that's what i got.

Answer 52

https://prnt.sc/phpg6y

Answer 53

dhyan kaha hai.?.?

Answer 54

U literally highlighted an empty space bro

Answer 55

empty??

Answer 56

Pakistan has lost 3 matches in a row bruhh... i’m already fuming over here... lol

Answer 57

highlighted transition y-->x

Answer 58

ohh really not following this time..

Answer 59

Oh my gook... one second plz

Answer 60

Oh no...I Isolated the dx/dt to the left side and dy/dt to the eight side

Answer 61

Oh no probably sounds like I made a mistake...but no...all I did was isolated the dx/dt the Rhs and dy/dt to the rhs

Answer 62

Oh my gook... I finally see it...

Answer 63

what i highlighted is a mistake..

Answer 64

You have laser eyes bruhh

Answer 65

Finally....you deserve women of the century award for helping me through this problem.... :)

Answer 66

ty.