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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0take 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
 5 years ago
Best ResponseYou've already chosen the best response.0ill do the one you posted so you can better understand

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, ok sorry about that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am not sure where i would find that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0here, tell me if this is the equation you are given: \[y ^{4}e ^{x} +x = y ^{6}x\] ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0Yes the equation is correct...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Its just applying the equation with the problem always gets me

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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
 5 years ago
Best ResponseYou've already chosen the best response.0so 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)+y61exy4 move the dy/dxs over 4y3ex(dy/dx)6y5x(dy/dx)=y61exy4 factor out dy/dx (dy/dx)(4y3ex6y5x)=y61exy4 solve for dy/dx by dividing the polynomial dy/dx=(y61exy4)/(4y3ex6y5x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that makes sense, thank you so much...i will apply that same rule to all the problems :)
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