Question

how to find the equation of a tangent line to the graph x^3 + y^3=6xy

Answer 1

First find the derivative with respect to x.

Answer 2

and that is
\[3x ^{2}+3y ^{2}=1\]

Answer 3

or do i differentiate expricitly?

Answer 4

You have to do it implicitly.

Answer 5

but i still didnt do it right?

Answer 6

No you didn't yet.

Answer 7

\[\left(\begin{matrix}d \\ dx\end{matrix}\right)\left[ x ^{3}+y ^{3}\right]=\left(\begin{matrix}d \\ dx\end{matrix}\right)\left[ 6xy \right]\]

Answer 8

so \[\left(\begin{matrix}dy \\ dx\end{matrix}\right)= \frac{ 3x ^{2} }{ -2x ^{2}}\]

Answer 9

do you know about chain rule ?

d/dx (y^3) = .... ?
(requires chain rule)

Answer 10

ugh okay so then do i do the chain rule from the very beggining??

Answer 11

d/dx(x^3) does not require chain rule

d/dx(6xy) requires chain and product rule

Answer 12

well what if i just moved the 6x?

Answer 13

moved where ?

keeping it as 6xy and applying chain+product rule is the easiest thing you can do here

Answer 14

what do u get for
d/dx (y^3) =... ?

Answer 15

3y

Answer 16

well
\[3y ^{2}\]

Answer 17

d/dy(y^3) = 3y^2

but here we are trying to find
d/dx(y^3)
notice the difference ?

Answer 18

so what am i doing then is it just 3y dy/dx?

Answer 19

if you mean \(3y^2 \dfrac{dy}{dx}\) , then yes!

what about 6xy ?
know product rule?

Answer 20

so what is f(x) and what is g(x) if there are three different variables instead of two?
or is f(x) 6x and g(x) y

Answer 21

6 is kust a constant, you can throw it out of the differentiation...

d/dx (6xy) = 6 d/dx(xy)

try to find
d/dx(xy)

Answer 22

isnt it just 1?

Answer 23

wait no never mind

Answer 24

x+y

Answer 25

product rule,
\((fg)' = f'g+fg'\)

\((xy)' = x'y+xy' \)

whats x' = ... ?

Answer 26

1

Answer 27

yes,
so,
6 d/dx(xy) is just 6 (y+ x dy/dx)
got this ?

Answer 28

but isnt the rule f(x)g'(x)+g(x)f'(x)??
so then it would be x(1)+y(1) therefore x+y??

Answer 29

\(\large y' = \dfrac{dy}{dx}\)

because we are differentiating with respect to y

Answer 30

i mean we are differentiating with respect to x ***

Answer 31

so then the dy/dx goes next to the y right?

Answer 32

yes,
so after your differentiation implicitly, you get
\(\Large 3x^2+3y^2 \dfrac{dy}{dx} = 6 (x\dfrac{dy}{dx}+y)\)

any doubts till here ?

Answer 33

i have that

Answer 34

so then i need to isolate the dy/dx on one side and factor it out?

Answer 35

yes!

Answer 36

how can i do that? do i divide by 6 first?

Answer 37

dividing by 6 will give you fractions...
lets do , divide by 3 on both sides.

Answer 38

then you can move y^2 dy/dx from left to right

Answer 39

why 3??

Answer 40

to avoid fractions...
if you divide by 6, on left side you get first term as
(1/2) x^2

Answer 41

why dont i destribute the 6?

Answer 42

you surely can do that...
and in later steps, you will realize that you need to divide 3 on both sides...
if you do it first or later, it will make no difference...

Answer 43

okay so after the three division?

Answer 44

\(\Large x^2+y^2 \dfrac{dy}{dx} = 2 (x\dfrac{dy}{dx}+y)\)

distribute the 2

Answer 45

then subtract 2y?

Answer 46

yes
what do u get ?

Answer 47

\[x ^{2}+y ^{2}\left(\begin{matrix}dy \\ dx\end{matrix}\right)-2y=2x \left(\begin{matrix}dy \\ dx\end{matrix}\right)\]

Answer 48

now to combine terms with dy/dx
you can subtract y^2 dy/dx from both sides,
try it

Answer 49

\[x ^{2}-2y=\left(\begin{matrix}dy \\ dx\end{matrix}\right)\left( 2xy ^{2} \right)\]

Answer 50

and then divide by 2xy^2

Answer 51

nooooooooooooo

Answer 52

why not

Answer 53

when you subtract, it'll be
\(\Large 2x-y^2\)
on right side

Answer 54

and not 2xy^2

Answer 55

wait why am i subtracting y^2?

Answer 56

isnt it multiplied by 2x

Answer 57

to combine terms with dy/dx together
\(x ^{2}+y ^{2}\left(\begin{matrix}dy \\ dx\end{matrix}\right)-2y=2x \left(\begin{matrix}dy \\ dx\end{matrix}\right) \\ x ^{2}+y ^{2}\left(\begin{matrix}dy \\ dx\end{matrix}\right)-y ^{2}\left(\begin{matrix}dy \\ dx\end{matrix}\right)-2y=2x \left(\begin{matrix}dy \\ dx\end{matrix}\right) -y ^{2}\left(\begin{matrix}dy \\ dx\end{matrix}\right)\\ x ^{2}-2y=(2x-y^2) \left(\begin{matrix}dy \\ dx\end{matrix}\right)\)

see if u get this ?

Answer 58

in the last step , i factored out dy.dx
so what remained was
2x and -y^2
so,
(2x-y^2)

Answer 59

alright i see what you did... and noooow you divide by 2x-y^2?

Answer 60

thats absolutely correct!
and you will get your required dy/dx :)

Answer 61

and that is the slope to my tangent line?

Answer 62

are you sure? because the slope should = -1

Answer 63

at which point do you need the slope ?

Answer 64

(3,3)

Answer 65

slope at 3,3 = dy/dx at 3,3

so put x= 3 and y=3 in the derivative dy/dx which you got

Answer 66

i think it would have been easier if we distributed the 6x first because then i could have moved dy/dx 6x to the left and from the left taken 3x and put that on the right

Answer 67

then i would have all the dy/dx on one side and factor it out and divide by that on the right and simplify into the -1/2

Answer 68

well -1

Answer 69

whatever you do,
if you get
dy/dx = (x^2-2y)/ (2x-y^2)
then you'll be correct, else incorrect.

Answer 70

and good, you got the slope as m=-1 as required :)

Answer 71

i wish i could show you the work.... i dont know why i didnt see it sooner

Answer 72

anyways thanks for your help and being patient and stuff.... sorry for taking up so much of your time

Answer 73

hey, no problem :)
always happy to help :)
welcome ^_^

Answer 74

i ended with dy/dx= 6y-3x/ 3y-6x and it simplifies down to negative one

Answer 75

oh, haha...now you know the correct dy/dx right ? :)
ask if u have any more doubts on this ?

Answer 76

i know the correct answer and all but i have a test on derivatives, rates of change, and related rates and i still dont understand it all and the test is tomorrow so im not very confident about it

Answer 77

don't worry much...practice 2 or 3 questions on each topic, and you'll be good to go :)

Answer 78

but the test is tomorrow and im already really stressed because im baking a pie

Answer 79

for pi day tomorrow

Answer 80

A real math person ^^^.

Answer 81

then why do i suck at math so much

Answer 82

i remember i used to like math and now im in calculus ksdjfihafiunoisdf

Answer 83

You just have to find how to understand the basics in a way that suits you; then you'll be golden.