Question

Write an equation for the tangent line to the graph of the following equation at the point (2,1):

Answer 1

\[x^3 +y^3 = 4xy+1\]

Answer 2

Did you find the derivative?

Answer 3

I had a little trouble with that...

Answer 4

What is the derivative of \(x^3\) ?

Answer 5

3x^2

Answer 6

yes, now how about y^3?
(Apply the same rule, but multiply times y')

Answer 7

I have a calculator that I can do all this on, but it ends up being a seriously detrimental crutch, so figuring out how to work it out is paramount. (<---- maybe some wordplay)

Answer 8

3y^2 * y'

Answer 9

yes, that is true

Answer 10

Okay, now apply the product rule to 4xy

Answer 11

what is the derivative of y?
what is the derivative of 4x?

Answer 12

It would be... 0? I feel like I am doing something wrong. Product rule is d/dx[f(x)g(x)]= f(x)g'(x) + g(x)f'(x)?

Answer 13

Derivative of 4x would be four actually, not zero, because the integer just 'comes along for the ride' as it were.

Answer 14

Yes, go on...

Answer 15

And y would be 1 x y or y x y' or 1 x y'? A bit weird there...

Answer 16

In general, the derivative of f(x) is f'(x) right?

Answer 17

Yes, f'(x) = the derivative of x

Answer 18

No, f'(x) is the derivative of the function f(x).

Answer 19

right?

Answer 20

In general, all I am asking is that you know that for any function f(x), the derivative is f'(x)... correct?

Answer 21

Yes

Answer 22

And now, you have to recall that y is also a function of x.
So the derivative of y is y'

Answer 23

Yes

Answer 24

And this way, when you differentiate 4x•y, you get:
(4)•y + 4x•(y')

Answer 25

Am I correct?

Answer 26

Sure, that makes sense.

Answer 27

Ok, now, when you differeniate each component you get:

\(\large\color{black}{ \displaystyle x^3 +y^3 = 4xy+1 }\)

\(\large\color{black}{ \displaystyle 3x^2 +3y^2y' = 4y+4xy'+0 }\)

Answer 28

May I ask are you finding the derivative with respect to y?

Answer 29

No, dy/dx.

Answer 30

Ok..watching and relearning

Answer 31

It is a derivative with respect to x.
Each y is a function of x, thus it deserves a chain rule of y'.

Answer 32

jmartine

Do you see how I get the equation below?

\(\large\color{black}{ \displaystyle 3x^2 +3y^2y' = 4y+4xy'+0 }\)

Answer 33

Yes, I do. "It is a derivative with respect to x.
Each y is a function of x, thus it deserves a chain rule of y'." That makes a lot of sense

Answer 34

Alright, now just using algebra you have to isolate the y'

Answer 35

3x^2+3y^2y′=4y+4xy′+0
3x^2-4y=3y^2y' + 4xy'
(3x^2-4y)/(3y^2 *4x)=2y'
(3x^2-4y)/(3y^2 *4x*2)=y'

Answer 36

So it would be \[y' = \frac{ 3x^2-4y }{ 3y^2 *4x*2}\]

Answer 37

\(\large\color{black}{ \displaystyle \frac{3x^2 -4y}{{4x-3y^2}}=y' }\)

Answer 38

Ok, I got a little messed up moving stuff from either side. I thought if you had
2x + 3x= 1 and you divided out 2 and 3 it would leave x + x or 2x? Maybe just me going blank on some core math stuff randomly...

Answer 39

Well, just a matter of algebraic manipulation...

Answer 40

\(\large\color{black}{ \displaystyle y'=\frac{3x^2 -4y}{4x-3y^2} }\)

gives you the expression for the function's slope.
That means that at (2,1) the slope is:

\(\large\color{black}{ \displaystyle y'=\frac{3(2)^2 -4(1)}{4(2)-3(1)^2}=\frac{12 -4}{8-3} =\frac{8}{5} }\)

Answer 41

So your line becomes

\(\large\color{black}{ \displaystyle y-y_1=m(x-x_1) }\)

\(\large\color{black}{ \displaystyle y-1=\frac{8}{5}(x-2) }\)

Answer 42

Ok, that definitely makes sense!

Answer 43

:)

Answer 44

Thanks for helping, I really understand how the whole process of solving the problem works now!