Question

Use implicit differentiation to find the slope of the tangent line to the curve

Answer 1

\[\frac{ y }{ x+4y} = x^8 - 9 \] at the point (1, -8/33)

Answer 2

So I am going to use to the quotient rule on the first term

Answer 3

\[\frac{ (x+4y)*y' - (y)(1+4y*y') }{ (x+4y)^2 }\]

Answer 4

that is the first term, can I simplify that?

Answer 5

it looks good
you can distribute and start gathering your y' terms on top

Answer 6

@freckles will it look like this: \[\frac{ y'x+4 * yy'-y+4y^2 * yy' }{( x + 4y)^2}\]

Answer 7

close

Answer 8

\[\frac{ y'x+4 * yy'-y-4y^2 * y' }{( x + 4y)^2}\]

Answer 9

okay, is that as far as I can simplify?

Answer 10

you can gather your y' terms if you want

Answer 11

you will need to do that to solve for y' later

Answer 12

have you differentiated the other side yet?

Answer 13

so \[\frac{ y'x + 4*yy' - y -4y2 * y' }{ (x + 4y)^2 } = 8x^7 - 0\]

Answer 14

ok that is beautiful so far

Answer 15

so the only thing to do is put your y'

Answer 16

isolate your y'

Answer 17

you could undo that ugly division there on the left hand side by multiplying both sides by the (x+4y)^2

Answer 18

and then gather those y's on the left hand side and put everything that doesn't have a y' on the opposite side

Answer 19

okay, let me give that a try

Answer 20

\[y = x^3 + 4x^2y - 9x -36y\]

Answer 21

now take the derivative of both sides?

Answer 22

\[y' = 3x^2 + 8xy'-9-36y'\]

Answer 23

whoa whoa

Answer 24

Wouldn't have used quotient rule. Would have used product rule.

Answer 25

you have already differentiated

Answer 26

something happened to all your y'

Answer 27

is this a different question you asking now

Answer 28

I'm confused

Answer 29

I am too

Answer 30

lol... I though you said to undo the division and multiply both sides

Answer 31

\[\frac{ y'x + 4*yy' - y -4y2 * y' }{ (x + 4y)^2 } = 8x^7 - 0 \\ y'x+4yy'-y-4y^2y'=8x^7(x+4y)^2 \]
didn't you have this

Answer 32

OH okay, I misunderstood, now let me see.

Answer 33

I multiply both sides by (x+4y)^2 to undo the division on the left

Answer 34

now you need to gather your terms on the left hand side that have y'

Answer 35

I am having the trouble figuring the proper way to gather those terms, distribution?

Answer 36

factoring

Answer 37

Since \( (x,y) =(1, -8/33)\)\[
\frac{ (x+4y)*y' - (y)(1+4y*y') }{ (x+4y)^2 }\\=
\frac{ ((1)+4(-8/33))*y' - (-8/33)(1+4(-8/33)*y') }{ ((1)+4(-8/33))^2 } = (1)^8-9
\]

Answer 38

like you know ab+ac+e
=a(b+c)+e

Answer 39

okay, let me try

Answer 40

and I think that -9 isn't suppose to be there and that should be 8(1)^7 on the right

Answer 41

\[y'(x + 4y)-y-4y^28y' = 8x^7\]

Answer 42

is the first term correct?

Answer 43

I forgot to append the rest on the second term

Answer 44

I think the 8 is a type-o
\[y'(x+4y-4y^2)-y=8x^7(x+4y)^2\]

Answer 45

now you are solving for y'
so add y on both sides
then divide both sides by what y' is multiply multiplied by

Answer 46

okay, let's see how messy this can get

Answer 47

it isn't too bad from here

Answer 48

just 2 steps to solve for y'

Answer 49

\[y' = \frac{8x^7(x+4y^2)+y }{ x + 4y - 4y^2}\]

Answer 50

I think the denominator can be factored

Answer 51

or maybe not... will it cancel out?

Answer 52

hmmm not sure I can play it with it ...

Answer 53

and I think that square is missplaced on top

Answer 54

\[y'=\frac{8x^7(x+4y)^2+y}{x+4y-4y^2}\]

Answer 55

I can't really see a prettier way to write it

Answer 56

maybe you can write in a different form

Answer 57

You do not need to find the implicit formula for \(y'\). You could have substituted in the point a long time ago to solve for the tangent line's slope.

Answer 58

okay, assuming its correct, let's find the slope of the tangent line. So plug the x in corrent?

Answer 59

@wio really?

Answer 60

Yes, look what I wrote above.

Answer 61

But that would be the slope is 8?

Answer 62

No, you would need to solve for \(y'\).

Answer 63

I am so lost.... I think I need to get rid of the Ys in the simplified equation

Answer 64

from what @wio had written (copy pasting) \[\frac{ (x+4y)*y' - (y)(1+4y*y') }{ (x+4y)^2 }\\=
\frac{ ((1)+4(-8/33))*y' - (-8/33)(1+4(-8/33)*y') }{ ((1)+4(-8/33))^2 } = (1)^8-9\]All you'd have to d is factor out y' from the left hand side and shift all the other stuff over. It makes finding y' a lot easier .

Answer 65

\[y=(x^8-9)(x+4y) \\ y'=(8x^7-0)(x+4y)+(x^8-9)(1+4y') \\ y'=8x^7(x+4y)+x^8+4x^8y'-36y'-9 \\ y'-4x^8y'+36y'=8x^7(x+4y)+x^8-9 \\ y' (1-4x^8+36)=8x^7(x+4y)+x^8-9 \\ y'=\frac{8x^7(x+4y)+x^8-9}{ 37-4x^8}\]
we could replace x^8-9 with y/(x+4y)
\[y'=\frac{8x^7(x+4y)+\frac{y}{x+4y}}{1-4\frac{y}{x+4y}} = \frac{8x^7(x+4y)^2+y}{(x+4y)-4y} =\frac{8x^7(x+4y)^2+y}{x}\]
So maybe we made a mistake somewhere
I would have to go back through it
--
you guys you are pluggin in the point without differentiating the right hand side

Answer 66

\[
\frac{ y }{ x+4y} = x^8 - 9
\] at the point \((1, -8/33)\)

I generally don't like the quotient rule, so I use the product rule.\[
\begin{split}
&y = (x^8-9)(x+4y) \\
\implies &y'=(x^8-9)'(x+4y)+(x^8-9)(x+4y)'\\
\implies &y'=8x^7(x+4y)+(x^8-9)(1+4y')
\end{split}
\]It is at this point I would sub in \((x,y) = (1,-8/33)\):\[
y'=8(1)^7(1+4(-8/33))+((1)^8-9)(1+4y')
\]Then it's a lot of simplification, and solving for \(y'\).

Answer 67

but about the y' at the end of that?

Answer 68

It's a variable. It will become easier to solve once you've gone through the order of operations.

Answer 69

The purpose of implicit differentiation is to make it easier to solve for a derivative.

Answer 70

@johnnydicamillo I see the mistake now

Answer 71

it was in the quotient rule earlier

Answer 72

\[\frac{ (x+4y)*y' - (y)(1+4y') }{ (x+4y)^2 }\]

Answer 73

so after making that correction you would have
\[y'=\frac{8x^7(x+4y)^2+y}{x+4y-4y} \]

Answer 74

since you already solved for y'
you could also put your point into that

Answer 75

and yes the 4y-4y is 0

Answer 76

either way though

Answer 77

okay to clear everything up \[y' = \frac{ (1)^7(1+4y)2+y }{ 1 }\]?

Answer 78

scratch that. typos

Answer 79

Suppose I even wanted to solve for \(y\) for whatever reason.
I shouldn't do it in this problem for my own time, but if I wanted to do so then...

\[
\frac{ y }{ x+4y} = x^8 - 9 \\
y = (x^8-9)(x+4y) =x(x^8-9)+4(x^8-9)y \\
y - 4(x^8-9)y =x(x^8-9)\\
y(1 - 4(x^8-9)) =x(x^8-9)\\
y = \frac{x(x^8-9)}{(1 - 4(x^8-9))}

\]

Answer 80

8/11?

Answer 81

\[y'=\frac{8(1)^7(1+4 \frac{-8}{33})^2+\frac{-8}{33}}{1}=8(1-\frac{32}{33})^2+\frac{-8}{33}=8(\frac{1}{33})^2-\frac{8}{33} =\frac{8}{33^2}-\frac{8}{33} \\ =\frac{8-8(33)}{33^2}\]
I think this is right
if i didn't make a mistake
and yes I know this is kinda ugly
I don't think that reduces to 8/11
looks negative to me
also you can check my work there
and finish it

Answer 82

-256/1089

Answer 83

yes!!!!!

Answer 84

You are the best! Super messy problem