A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
Use implicit differentiation to find the slope of the tangent line to the curve2xy^3+5xy=35 at the point (5,1).
anonymous
 one year ago
Use implicit differentiation to find the slope of the tangent line to the curve2xy^3+5xy=35 at the point (5,1).

This Question is Open

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Hey burns :) What are you having trouble with? We need to find y' at the specified point, ya?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0\[\Large\rm 2xy^3+5xy=35 \]Differentiating:\[\Large\rm \color{royalblue}{(2x)'}y^3+2x\color{royalblue}{(y^3)'}+\color{royalblue}{(5x)'}y+5x\color{royalblue}{(y)'}=\color{royalblue}{(35)'}\]We would have to apply the product rule in both cases, do you understand this setup?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How did you get that set up? i know the product rule, I am just unfimiliar with how to get it set up properly

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0To get more comfortable with product rule, I would recommend doing the "setup" as I did above, eventually you can skip that step when you feel good at these. Example:\[\Large\rm (xy)'=x'y+xy'\] If differentiating with respect to x, the x' becomes 1 simply,\[\Large\rm (xy')=1\cdot y+xy'\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For implicit differentiation, just treat x and y as regular variables for derivatives. For every y that you derive, you need to add a y' after derivation,

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0\[\Large\rm (\color{orangered}{2x}\color{royalblue}{y^3})'=(\color{orangered}{2x})'\color{royalblue}{y^3}+\color{orangered}{2x}(\color{royalblue}{y^3})'\]Ok with this setup for the first term? :o

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0You can group the 2 with either the x OR the y^3, it doesn't matter which, just be consistent.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0So after we have our setup, we apply our differentiation,\[\Large\rm (\color{orangered}{2x}\color{royalblue}{y^3})'=(2)\color{royalblue}{y^3}+\color{orangered}{2x}(3y^2y')\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Having any confusion as to why we have y' showing up, but not x'? :O

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is it because we canceled out the x's?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Ummmm, well it's because of the stuff that we end up with:\[\Large\rm \frac{d}{dx}x^2=2x\frac{d}{dx}x\]Here is a simple chain rule being applied ^ We differentiated the outermost function, and then we multiply by the derivative of the inner function,\[\Large\rm \frac{d}{dx}x^2=2x\frac{dx}{dx}\]But this thing really has no significance, that's why they usually ignore doing this chain when you're learning your power rule and such. It's like.. if you were to read into it, the numerator says "how much does x change" with the denominator "as x changes an immeasurably small amount?"\[\Large\rm \frac{d}{dx}x^2=2x\cdot 1\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0I dunno, maybe that's an unnecessary and longwinded explanation, i'm just trying to explain that we're always apply the chain rule in this way, but it only actually matters when it's a variable that doesn't match our operator variable.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0So by comparison,\[\Large\rm \frac{d}{dx}y^2=2y\frac{d}{dx}y\]giving us,\[\Large\rm \frac{d}{dx}y^2=2y\frac{dy}{dx}\]or\[\Large\rm \frac{d}{dx}y^2=2y y'\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0So just to review that first term again, setup,\[\Large\rm (2xy^3)'=(2x)'y^3+2x(y^3)'\]differentiate,\[\Large\rm (2xy^3)'=(2x')y^3+2x(3y^2y')\]and simplify,\[\Large\rm (2xy^3)'=(2)y^3+2x(3y^2y')\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and then plug in the points?
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.