Having a few problems needing help. Trying to Find dy/dx by implicit differentiation. y4ex + x = y6x

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Having a few problems needing help. Trying to Find dy/dx by implicit differentiation. y4ex + x = y6x

Mathematics
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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=______
ill do the one you posted so you can better understand
is y4 y^4?

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yes, ok sorry about that
and is ex e^x?
I am not sure where i would find that?
here, tell me if this is the equation you are given: \[y ^{4}e ^{x} +x = y ^{6}x\] ?
would dy=y 6 -y4ex-1
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?
Yes the equation is correct...
Its just applying the equation with the problem always gets me
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.
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)
that makes sense, thank you so much...i will apply that same rule to all the problems :)

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