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monroe17
Group Title
can someone help me step by step?
Find the derivative
3x^(2)+e^(2y)=4y+tan(x)
 one year ago
 one year ago
monroe17 Group Title
can someone help me step by step? Find the derivative 3x^(2)+e^(2y)=4y+tan(x)
 one year ago
 one year ago

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Operative Group TitleBest ResponseYou've already chosen the best response.0
solve for dx right?
 one year ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.2
Have you learned implicit differentiation?
 one year ago

monroe17 Group TitleBest ResponseYou've already chosen the best response.1
I'm not good at it.
 one year ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.2
It's easy! Let me explain it to you in simple terms without getting wordy. If x and y are on the same side of the equal sign, then we say the function is written implicitly, and call it an implicit function. To find the derivative of an implicit function, take the derivative as you normally would but every time you differentiate y, multiply by dy/dx. Then solve for dy/dx. Example: x² + y³ = 5 2x + 3y²(dy/dx) = 0 3y²(dy/dx) = 2x dy/dx = 2x/3y² Understand @monroe17 ?
 one year ago

monroe17 Group TitleBest ResponseYou've already chosen the best response.1
I got it :) and when you have both x and y on each side you move the y terms to the left? and x on the right then divide?
 one year ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.2
Yes you isolate dy/dx and then solve for dy/dx. Good, you got it!
 one year ago

monroe17 Group TitleBest ResponseYou've already chosen the best response.1
for instance with my problem i would subtract 6x from both sides and it ends up on the right side of the equal sign. Then 4 is substracted to the right side.. getting sec^2(x)6x/2e^(2y)4?
 one year ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.2
Excellent!!
 one year ago

monroe17 Group TitleBest ResponseYou've already chosen the best response.1
yay, gosh finally.. imma try it for my next problem and see if i get it right.
 one year ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.2
Sure! I'll be on line for another hour or so, hence if you need anything, let me know.
 one year ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.2
Welcome! Good Luck!
 one year ago

monroe17 Group TitleBest ResponseYou've already chosen the best response.1
opps. I came to a problem.. ;/ my next question is 5y^2=3x^(2)y+e^(x) i did.. 10y(dy/dx)=6x(dy/dx)+e^(x) dy/dx(10y*y)=6x+e^(x) dy/dx=6x+e^(x)/10y*y and that's wrong.. what was my mistake??
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Woops! When you took the derivative of the right side, it looks like you didn't apply the product rule on the 3x^2y term! :O
 one year ago

monroe17 Group TitleBest ResponseYou've already chosen the best response.1
ohhhh okay let me try that again.
 one year ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.2
Exactly as @zepdrix said!
 one year ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.2
@monroe17 remember you only multiply by dy/dx when you take the derivative of y, not every time you write y.
 one year ago

monroe17 Group TitleBest ResponseYou've already chosen the best response.1
oh i don't ?
 one year ago

monroe17 Group TitleBest ResponseYou've already chosen the best response.1
I got.. 10y(dy/dx)=3x^2+6x*y+e^(x)
 one year ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.2
Example: x² + 2xy  4 = 6y 2x + 2y + 2x(dy/dx) = 6(dy/dx) Do you see what I did there?
 one year ago

monroe17 Group TitleBest ResponseYou've already chosen the best response.1
how do i know which term goes to which side for this problem.
 one year ago

Eda2012 Group TitleBest ResponseYou've already chosen the best response.0
\[5y ^{2} = 3x ^{2}y + e ^{x}\] \[10 y \frac{ dy }{ dx} = 3x ^{2 }\frac{ dy }{ dx } + 6xy + e ^{x}\] \[\frac{ dy }{ dx } = \frac{ 6xy + e ^{x} }{ 10y  3x }\]
 one year ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.2
You know the product rule for derivatives, right?
 one year ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.2
@Eda2012 I am trying to teach her. I do not appreciate you just posting the solution. How is she supposed to learn to solve problems independently if you just give her the answers?
 one year ago

monroe17 Group TitleBest ResponseYou've already chosen the best response.1
^thank you! lol
 one year ago

monroe17 Group TitleBest ResponseYou've already chosen the best response.1
the product rule is f(x)g'(x)+g(x)f'(x) right?
 one year ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.2
Correct! Thus if you have a term of 3x²y for example, then your f = 3x² and g = y. And you multiply by dy/dx iff you differentiate y. Understand?
 one year ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.2
If f = 3x² and g = y, then f ' = 6x and g' = dy/dx Do you see?
 one year ago

monroe17 Group TitleBest ResponseYou've already chosen the best response.1
so.. 10y(dy/dx)=3x^2(dy/dx)+y(6x)+e^(x) 10y(dy/dx)=3x^(2)+6xy+e^(x)
 one year ago

monroe17 Group TitleBest ResponseYou've already chosen the best response.1
why does 3x^2 go to the left side? I though you make all x to one side and y's the other?
 one year ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.2
(fg)' = fg' + gf' (3x²y)' = 3x²(dy/dx) + 6xy Do you see your mistake when you differentiated 3x²y on the right side?
 one year ago

Eda2012 Group TitleBest ResponseYou've already chosen the best response.0
@calculusfunctions sometimes you have to show some works and then explain it...sorry...
 one year ago

monroe17 Group TitleBest ResponseYou've already chosen the best response.1
oh yeah my bad. I mixed them up.
 one year ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.2
@Eda2012 so you show examples and explain the lessons not just give out the answers. I am a teacher so you really don't need to tell me how to do my job!
 one year ago

Eda2012 Group TitleBest ResponseYou've already chosen the best response.0
@calculusfunctions ok...next time i won't do it again...sorry
 one year ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.2
@monroe17 do you understand now?
 one year ago

monroe17 Group TitleBest ResponseYou've already chosen the best response.1
^ calculus it's okay, I ignored it :) why does 3x^2 go to the left side? I though you make all x to one side and y's the other?
 one year ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.2
Because it is 3x²(dy/dx) and you're isolating dy/dx.
 one year ago

monroe17 Group TitleBest ResponseYou've already chosen the best response.1
... im confused
 one year ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.2
OK Do you want me to show you step by step?
 one year ago

monroe17 Group TitleBest ResponseYou've already chosen the best response.1
let me look at it real quick.
 one year ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.2
Since @eda2012 already posted the correct solution, go ahead and observe it and then tell me which of her steps confuses you, if any. Alright?
 one year ago

monroe17 Group TitleBest ResponseYou've already chosen the best response.1
okay, give me a little time
 one year ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.2
Sure! Someone else is asking for my help so I'll go and get them started. You call me when you're ready. Take your time.
 one year ago

monroe17 Group TitleBest ResponseYou've already chosen the best response.1
I'm sorry that took forever, I had a lab report to complete. But, I relooked over the problem and it seems to make a little more sense. Thank you so much for your help!
 one year ago
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