Question

Need help with implicit differentiation.

Consider the implicit function y*(xy+1)=4(x+4)
a) what is the implicit derrivative?
b)Find the tangent line passing through the point (1,4)
c) Does the graph have any horizontal lines? if so, where?

Any guidance on this problem would be greatly appreciated.

Answer 1

have you tried differentiating both sides?

Answer 2

You are applying the same rules for taking the derivative !

(Every time when you differentiate something that contains y you
will multiply times y\('\) -- this is because y is really a function of x,
and therefore it gets its own chain-rule, JUST AS a derivative of sin(4x),
\(\sqrt{x^3+4x-3}\), or many other functions would get.)

Answer 3

So I would just need to distribute, then find the derivative of the entire problem?

xy^2+y=4x+16

to 2xy * y' + 1*y'=4?

then to 2y'=4-2xy

to y'=(4-2xy)/2?

Answer 4

@SolomonZelman

Answer 5

hi

Answer 6

\[xy^2+y=4x+16 \\ \text{ you need to use product rule for the } (xy^2)' \text{ part }\]

Answer 7

\[\frac{d}{dx}(xy^2)= y^2 \cdot \frac{d}{dx}(x)+x \cdot \frac{d}{dx}(y^2) \\ =?\]

Answer 8

so after that, I found \[\frac{ dy }{ dx }=\frac{ 4-y^2 }{ 4xy },\] But I feel like I may have done something wrong.

Answer 9

Nevermind, I found my mistake. The derivative should actually be \[\frac{ dy }{ dx }=\frac{ 4-y^2 }{ 2xy+1 }\]

Answer 10

Then to find the tangent line at 1,4 I believe we would need to plug these x and y values into the derivative to give us our slope, so: \[\frac{ 4-(4)^2 }{ 2(1)(4)+1 } = \frac{ -12 }{ 9}\]

Answer 11

Using that in the formula y-y_1=m(x-x_1) would give us \[y=\frac{ -4 }{ 3 }(x-1)+4\]

Answer 12

So, I just need help finding if it has any horizontal tangents, and where they are located. I believe we would need to set our derivative equal to zero, but I dont know what to do now that there are an x and a y.

Answer 13

hey

Answer 14

Hello

Answer 15

\[xy^2+y=4x+16 \\ x \cdot 2y y'+y^2+y'=4 \\ y'(2xy+1)=4-y^2 \\ y'=\frac{4-y^2}{2xy+1}\]
your y' looks awesome!

Answer 16

\[y'|_{(1,4)}=\frac{4-4^2}{2(1)(4)+1}=\frac{-12}{9}=\frac{-4}{3} \\ y-f(1)=f'(1)(x-1) \\ y-4=\frac{-4}{3}(x-1)\]
your tangent line at (1,4) looks great

so now you are looking for horizontal tangents you say

Answer 17

horizontal tangents occur when y'=0

Answer 18

y' we have written as a fraction
fractions are 0 when the numerator is 0

Answer 19

\[xy^2+y=4x+16 \\ x \cdot 2y y'+y^2+y'=4 \\ y'(2xy+1)=4-y^2 \\ y'=\frac{4-y^2}{2xy+1} \\ 4-y^2=0\]

Answer 20

4-y^2=0 this equations is going to help us find the horizontal tangents

Answer 21

okay :)

Answer 22

notice this equation is just in terms of y
so we solve for y
and find x by pluggin into the original equation

Answer 23

So y would be y= +/- 2?

Answer 24

right
now plug into original function to find the corresponding x values

Answer 25

hey and also when we find these x values...

Answer 26

we want to make sure it doesn't make y' undefined when pluggin in the point
like we do not want the 1+2xy to be 0

Answer 27

anyways let's see what happens

Answer 28

\[x(2)^2+(2)=4x+16\]
\[4x+2=4x+16\]

since these can't be equal, could we assume that there are no tangent lines?

Answer 29

oops I replace the x with 2

Answer 30

I thought that the y was 2?

Answer 31

yeah that is why I said oops :)
\[xy^2+y=4x+16 \\ y=2 \\ 4x+4=4x+16 \\ y=-2 \\ 4x-2=4x+16 \]
there is no x for either of trhese

Answer 32

so there are no horizontal tangents

Answer 33

ooh, haha I thought you wanted me to put x as 2, sorry about that!

Answer 34

Okay, thank you so much!

Answer 35

nope it was my oops not your oops :)

Answer 36

So to find this tangent line, we took the top part of the derivative. If we wanted to find vertical lines, would we set the bottom equal to 0?

Answer 37

yep
vertical lines can happen when the denominator of y' is 0

Answer 38

ooh, so having the top of the fraction equal to 0 would make the entire thing 0, but with a 0 in the denominator, it would be undefined, making it vartical.

Answer 39

Vertical*

Answer 40

well it could make it vertical

Answer 41

it could be just no continuous there
and that is y' doesn't exist

Answer 42

oh yeah, thats a good point

Answer 43

\[xy^2+y=4x+16 \\ y'=\frac{4-y^2}{2xy+1} \\ 2xy+1 =0 \\ xy=\frac{-1}{2} \\\\ xy^2+y=4x+16 \\ \text{ multiply } x \text{ on both sides } \\ x^2y^2+xy=4x^2+16x \\ \text{ replace } xy \text{ with } \frac{-1}{2} \\ (\frac{-1}{2})^2+\frac{-1}{2}=4x^2+16x \\ \]
\[\frac{1}{4}-\frac{1}{2}=4x^2+16 x \\ 1-2=16x^2+64x \\ 16x^2+64x+1=0\]

Answer 44

What does that final part tell us?

Answer 45

\[x=-2 \pm \frac{3 \sqrt{7}}{4} \\ x=-2+\frac{3 \sqrt{7}}{4} \approx -0.0157 \\ x=-2-\frac{3 \sqrt{7}}{4} \approx -3.984\]
to see if this point actually exist we would have to plug in this x into original to see if there is a corresponding y that exists
http://www.wolframalpha.com/input/?i=%28-2%2B3+sqrt%287%29%2F4%29y%5E2%2By%3D4%28-2%2B3+sqrt%287%29%2F4%29%2B16
\[\text{ this above link says when } x=-2+\frac{3 \sqrt{7}}{4} \text{ that we have } y=16+6 \sqrt{7} \approx 31.875 \\ \\\]

http://www.wolframalpha.com/input/?i=%28-2-3+sqrt%287%29%2F4%29y%5E2%2By%3D4%28-2-3+sqrt%287%29%2F4%29%2B16
\[\text{ this above link says when } x=-2+\frac{3 \sqrt{7}}{4} \text{ that we have } y=16-6 \sqrt{7} \approx 0.12549 \\ \\\]

hmm... I wonder if xy=-1/2...

Answer 46

yep just verified that last sentence

Answer 47

Since theres an actual point in both, wold this imply no vertical slopes either?

Answer 48

yep
we have horizontal tangents when the y' numerator is 0 but its denominator isn't 0

we have vertical tangents when y' denominator is 0 but its numerator isn't 0

Answer 49

Awesome, thank you!

Answer 50

http://www.wolframalpha.com/input/?i=+xy%5E2%2By%3D4x%2B16
I don't get wolfram's graph

Answer 51

If I zoom maybe I see a vertical tangent close to (-0.015,31)
http://www.wolframalpha.com/input/?i=+xy%5E2%2By%3D4x%2B16%2Cx+in+%28-1%2C0%29

Answer 52

I just realized I made a type-o above
\[\text{ this above link says when } x=-2-\frac{3 \sqrt{7}}{4} \text{ that we have } y=16-6 \sqrt{7} \approx 0.12549 \\ \\\]

Answer 53

http://www.wolframalpha.com/input/?i=+xy%5E2%2By%3D4x%2B16%2Cx+in+%28-4%2C-3%29%2Cy+in+%280%2C.5%29
for the one that is close to (-3.98,.12)
I guess I just had to do a lot of zooming in

Answer 54

@freckles Even though there was no x values that made 4x+2=4x+16?

Answer 55

what is this question about?

Answer 56

The graph you put up showed that there was a vertical tangent line at (-3.98,.12), but when we tried to solve it algebraically, it showed that there wasnt.

Answer 57

no

Answer 58

I think you are getting horizontal and vertical mixed up

Answer 59

we had no horizontal tangents because 4x+4=4x+16 or 4x-2=4x+16 didn't have a solution
but we did find two vertical tangents

Answer 60

oooh, yeah, never mind