anonymous
  • anonymous
How do you implicitly differentiate: x^3y^3-y=x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
take y to the other side then apply product rule in the left hand side
anonymous
  • anonymous
gimme a sec i am working on it
amistre64
  • amistre64
you aint gots to move anything, just differentiate

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More answers

anonymous
  • anonymous
haha amistre is correct
amistre64
  • amistre64
:) it happens
amistre64
  • amistre64
x^3y^3-y=x x'^3y^3+x^3y^'3-y'=x'
amistre64
  • amistre64
eventually you factor out a y' and move it all to the other side tho
anonymous
  • anonymous
d/dx(x^3y^3)-d/dx(y)=d/dx(x)
anonymous
  • anonymous
For d/dx(x^3y^3) you use the Chain Rule?
amistre64
  • amistre64
product rule, and some say chain rule for the y(x) part
anonymous
  • anonymous
What are the steps please?
anonymous
  • anonymous
\[x^3y^3-y=x\] is like \[x^3f(x)^3-f(x)=x\]
anonymous
  • anonymous
for first term you use the product rule and the chain rule \[3x^2f^3(x)+x^33f^2(x)f'(x)\] or \[3x^2y^3+3x^3y^2y'\]
anonymous
  • anonymous
so start with \[3x^2y^3+3x^3y^2y-y'=1\] then solve for \[y'\]
anonymous
  • anonymous
rather \[3x^2y^3+3x^3y^2y'-y'=1\]
anonymous
  • anonymous
So the answer is: \[y' = (1-3x^2y^3)/(3x^2-1)\]
amistre64
  • amistre64
forgot a y^2 in the denom
amistre64
  • amistre64
but other than that, yes
anonymous
  • anonymous
So it's \[y′=(1−3x^2y^3)/(3x^2−y^2)\]
amistre64
  • amistre64
\[y′=\frac{1−3x^2y^3}{3x^2y^2−1}\]
anonymous
  • anonymous
I'm trying to figure out where I forgot the y^2..
amistre64
  • amistre64
\[3x^2y^3+3x^3y^2y'-y'=1\] \[3x^3y^2y'-y'=1-3x^2y^3\] \[y'(3x^3y^2-1) =1-3x^2y^3\]
anonymous
  • anonymous
Yes, the correct one is: y' = ( 1- 3x^2 y^3 ) / ( 3x^3 y^2 -1)

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