anonymous
  • anonymous
i have some questions about implicit differentiation
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[xy(2x+3y)=2\]
anonymous
  • anonymous
ok my question is do i need to set up the multiplication derivative equation for both xy(2x+3y) as well as xy?
myininaya
  • myininaya
You could distribute there!

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anonymous
  • anonymous
ok so i should distribute first makes sense
myininaya
  • myininaya
Well it will make your life a lot easier
anonymous
  • anonymous
so know i have \[2x^2y+2xY^2=2 \] so know i am going to have to differentiate using the product equation for both 2x^2y and 2xy^2 right?
anonymous
  • anonymous
sorry that should be 3xy^2
myininaya
  • myininaya
Yes since you have a product for both terms you use the product rule for both terms
myininaya
  • myininaya
You might consider some of these useful for your problem: \[(fg)'=f'g+fg' ; (cfg)'=c(fg)'; (y^n)'=ny^{n-1}y' ; (x^n)'=nx^{n-1}\]
myininaya
  • myininaya
\[(3xy^2)'=3(xy^2)' \text{ by constant multiple rule}\] \[=3 \cdot [(x)'y^2+x(y^2)'] \text{ by product rule}\] So we know (x)'=1 So we have: \[=3[1y^2+x(2y^{2-1}y')] \text{ by chain rule}\] So finally we can say all nice like that \[(3xy^2)'=3[y^2+2xyy']\]
myininaya
  • myininaya
\[(3xy^2)'=3y^2+6xyy'\]
myininaya
  • myininaya
Do you see how I applied the rules?
myininaya
  • myininaya
You try the other product and I will check k?
anonymous
  • anonymous
ok
myininaya
  • myininaya
This one: \[(3x^2y)'\] And remember that y=f(x) :means y is a function of x
anonymous
  • anonymous
ok i got 4xy+2x^2y^prime
anonymous
  • anonymous
and i used (2x^2y)prime
myininaya
  • myininaya
Ok! \[(2x^2y)'=2(x^2y)'=2(2xy+x^2y')=4xy+2x^2y'\] Beautiful! :)
anonymous
  • anonymous
sweet! thanks a ton!
myininaya
  • myininaya
np! :)

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