anonymous
  • anonymous
Having a few problems needing help. Trying to Find dy/dx by implicit differentiation. y4ex + x = y6x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
take the derivative of each part as though it was a typical derivative, except that every time you take the derivative of y, you multiply that part by dy/dx, then once its all differentiated, put all the parts multiplied by dy/dx on one side of the equation and the non dy/dxs on the other side of the equal sign. then simplify so that you have dydx=______
anonymous
  • anonymous
ill do the one you posted so you can better understand
anonymous
  • anonymous
is y4 y^4?

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anonymous
  • anonymous
??
anonymous
  • anonymous
yes, ok sorry about that
anonymous
  • anonymous
and is ex e^x?
anonymous
  • anonymous
I am not sure where i would find that?
anonymous
  • anonymous
here, tell me if this is the equation you are given: \[y ^{4}e ^{x} +x = y ^{6}x\] ?
anonymous
  • anonymous
would dy=y 6 -y4ex-1
anonymous
  • anonymous
i see where your going, but that wont work in this case. an easier way would be to just go through it, is the equation i wrote above correct?
anonymous
  • anonymous
Yes the equation is correct...
anonymous
  • anonymous
Its just applying the equation with the problem always gets me
anonymous
  • anonymous
ok so then, and this applies to all equations, when you are using implicit differentiation, you can just do the following: y4ex+x=y6x (4y3ex(dy/dx)+exy4)+1=6y5x(dy/dx)+y6 i used the product rule twice there, and whenever i took the derivative of y, i multiplied it by "dy/dx" because we are differentiating in terms of x.
anonymous
  • anonymous
so then you take that (4y3ex(dy/dx)+exy4)+1=6y5x(dy/dx)+y6 = 4y3ex(dy/dx)+exy4+1=6y5x(dy/dx)+y6 move the non dy/dx over 4y3ex(dy/dx)=6y5x(dy/dx)+y6-1-exy4 move the dy/dxs over 4y3ex(dy/dx)-6y5x(dy/dx)=y6-1-exy4 factor out dy/dx (dy/dx)(4y3ex-6y5x)=y6-1-exy4 solve for dy/dx by dividing the polynomial dy/dx=(y6-1-exy4)/(4y3ex-6y5x)
anonymous
  • anonymous
that makes sense, thank you so much...i will apply that same rule to all the problems :)

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