Revircs
  • Revircs
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.
Calculus1
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
freckles
  • freckles
have you tried differentiating both sides?
SolomonZelman
  • SolomonZelman
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.)
Revircs
  • Revircs
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?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Revircs
  • Revircs
@SolomonZelman
freckles
  • freckles
hi
freckles
  • freckles
\[xy^2+y=4x+16 \\ \text{ you need to use product rule for the } (xy^2)' \text{ part }\]
freckles
  • freckles
\[\frac{d}{dx}(xy^2)= y^2 \cdot \frac{d}{dx}(x)+x \cdot \frac{d}{dx}(y^2) \\ =?\]
Revircs
  • Revircs
so after that, I found \[\frac{ dy }{ dx }=\frac{ 4-y^2 }{ 4xy },\] But I feel like I may have done something wrong.
Revircs
  • Revircs
Nevermind, I found my mistake. The derivative should actually be \[\frac{ dy }{ dx }=\frac{ 4-y^2 }{ 2xy+1 }\]
Revircs
  • Revircs
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: \[\frac{ 4-(4)^2 }{ 2(1)(4)+1 } = \frac{ -12 }{ 9}\]
Revircs
  • Revircs
Using that in the formula y-y_1=m(x-x_1) would give us \[y=\frac{ -4 }{ 3 }(x-1)+4\]
Revircs
  • Revircs
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.
freckles
  • freckles
hey
Revircs
  • Revircs
Hello
freckles
  • freckles
\[xy^2+y=4x+16 \\ x \cdot 2y y'+y^2+y'=4 \\ y'(2xy+1)=4-y^2 \\ y'=\frac{4-y^2}{2xy+1}\] your y' looks awesome!
freckles
  • freckles
\[y'|_{(1,4)}=\frac{4-4^2}{2(1)(4)+1}=\frac{-12}{9}=\frac{-4}{3} \\ y-f(1)=f'(1)(x-1) \\ y-4=\frac{-4}{3}(x-1)\] your tangent line at (1,4) looks great so now you are looking for horizontal tangents you say
freckles
  • freckles
horizontal tangents occur when y'=0
freckles
  • freckles
y' we have written as a fraction fractions are 0 when the numerator is 0
freckles
  • freckles
\[xy^2+y=4x+16 \\ x \cdot 2y y'+y^2+y'=4 \\ y'(2xy+1)=4-y^2 \\ y'=\frac{4-y^2}{2xy+1} \\ 4-y^2=0\]
freckles
  • freckles
4-y^2=0 this equations is going to help us find the horizontal tangents
Revircs
  • Revircs
okay :)
freckles
  • freckles
notice this equation is just in terms of y so we solve for y and find x by pluggin into the original equation
Revircs
  • Revircs
So y would be y= +/- 2?
freckles
  • freckles
right now plug into original function to find the corresponding x values
freckles
  • freckles
hey and also when we find these x values...
freckles
  • freckles
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
freckles
  • freckles
anyways let's see what happens
Revircs
  • Revircs
\[x(2)^2+(2)=4x+16\] \[4x+2=4x+16\] since these can't be equal, could we assume that there are no tangent lines?
freckles
  • freckles
oops I replace the x with 2
Revircs
  • Revircs
I thought that the y was 2?
freckles
  • freckles
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
freckles
  • freckles
so there are no horizontal tangents
Revircs
  • Revircs
ooh, haha I thought you wanted me to put x as 2, sorry about that!
Revircs
  • Revircs
Okay, thank you so much!
freckles
  • freckles
nope it was my oops not your oops :)
Revircs
  • Revircs
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?
freckles
  • freckles
yep vertical lines can happen when the denominator of y' is 0
Revircs
  • Revircs
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.
Revircs
  • Revircs
Vertical*
freckles
  • freckles
well it could make it vertical
freckles
  • freckles
it could be just no continuous there and that is y' doesn't exist
Revircs
  • Revircs
oh yeah, thats a good point
freckles
  • freckles
\[xy^2+y=4x+16 \\ y'=\frac{4-y^2}{2xy+1} \\ 2xy+1 =0 \\ xy=\frac{-1}{2} \\\\ xy^2+y=4x+16 \\ \text{ multiply } x \text{ on both sides } \\ x^2y^2+xy=4x^2+16x \\ \text{ replace } xy \text{ with } \frac{-1}{2} \\ (\frac{-1}{2})^2+\frac{-1}{2}=4x^2+16x \\ \] \[\frac{1}{4}-\frac{1}{2}=4x^2+16 x \\ 1-2=16x^2+64x \\ 16x^2+64x+1=0\]
Revircs
  • Revircs
What does that final part tell us?
freckles
  • freckles
\[x=-2 \pm \frac{3 \sqrt{7}}{4} \\ x=-2+\frac{3 \sqrt{7}}{4} \approx -0.0157 \\ x=-2-\frac{3 \sqrt{7}}{4} \approx -3.984\] 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...
freckles
  • freckles
yep just verified that last sentence
Revircs
  • Revircs
Since theres an actual point in both, wold this imply no vertical slopes either?
freckles
  • freckles
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
Revircs
  • Revircs
Awesome, thank you!
freckles
  • freckles
http://www.wolframalpha.com/input/?i=+xy%5E2%2By%3D4x%2B16 I don't get wolfram's graph
freckles
  • freckles
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
freckles
  • freckles
I just realized I made a type-o above \[\text{ this above link says when } x=-2-\frac{3 \sqrt{7}}{4} \text{ that we have } y=16-6 \sqrt{7} \approx 0.12549 \\ \\\]
freckles
  • freckles
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
Revircs
  • Revircs
@freckles Even though there was no x values that made 4x+2=4x+16?
freckles
  • freckles
what is this question about?
Revircs
  • Revircs
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.
freckles
  • freckles
no
freckles
  • freckles
I think you are getting horizontal and vertical mixed up
freckles
  • freckles
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
Revircs
  • Revircs
oooh, yeah, never mind

Looking for something else?

Not the answer you are looking for? Search for more explanations.