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monroe17
can someone help me step by step? Find the derivative 3x^(2)+e^(2y)=4y+tan(x)
Have you learned implicit differentiation?
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 ?
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?
Yes you isolate dy/dx and then solve for dy/dx. Good, you got it!
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?
yay, gosh finally.. imma try it for my next problem and see if i get it right.
Sure! I'll be on line for another hour or so, hence if you need anything, let me know.
Welcome! Good Luck!
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??
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
ohhhh okay let me try that again.
Exactly as @zepdrix said!
@monroe17 remember you only multiply by dy/dx when you take the derivative of y, not every time you write y.
I got.. 10y(dy/dx)=3x^2+6x*y+e^(x)
Example: x² + 2xy - 4 = 6y 2x + 2y + 2x(dy/dx) = 6(dy/dx) Do you see what I did there?
how do i know which term goes to which side for this problem.
\[5y ^{2} = 3x ^{2}y + e ^{x}\] \[10 y \frac{ dy }{ dx} = 3x ^{2 }\frac{ dy }{ dx } + 6xy + e ^{x}\] \[\frac{ dy }{ dx } = \frac{ 6xy + e ^{x} }{ 10y - 3x }\]
You know the product rule for derivatives, right?
@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?
the product rule is f(x)g'(x)+g(x)f'(x) right?
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?
If f = 3x² and g = y, then f ' = 6x and g' = dy/dx Do you see?
so.. 10y(dy/dx)=3x^2(dy/dx)+y(6x)+e^(x) 10y(dy/dx)=3x^(2)+6xy+e^(x)
why does 3x^2 go to the left side? I though you make all x to one side and y's the other?
(fg)' = fg' + gf' (3x²y)' = 3x²(dy/dx) + 6xy Do you see your mistake when you differentiated 3x²y on the right side?
@calculusfunctions sometimes you have to show some works and then explain it...sorry...
oh yeah my bad. I mixed them up.
@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!
@calculusfunctions ok...next time i won't do it again...sorry
@monroe17 do you understand now?
^ 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?
Because it is 3x²(dy/dx) and you're isolating dy/dx.
OK Do you want me to show you step by step?
let me look at it real quick.
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?
okay, give me a little time
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.
I'm sorry that took forever, I had a lab report to complete. But, I re-looked over the problem and it seems to make a little more sense. Thank you so much for your help!