Calculus

Derivative of the implicit Function:

arctan(x^5y)=xy^5

Here is where I am at. I am pretty frustrated:

arctan(x^5y)-xy^5=0

1/1+(x^5y)^2 * (d/dx(x^5)*y+x^5*d/dx(y))-(d/dx(x)*y^5+x*d/dx(y^5))=0

1/(1+x^5y)^2 * (5x^4y+x^5dy/dx) - (y^5+x^5y^4dy/dx)

(5x^4y)/[(1+x^5y)^2] + (x^5dy/dx)/[(1+x^5y)^2] - y^5+5y^4dy/dx=0

I move the parts of the problem that lack the dy/dx to the other side of the equation for

[X^5 (dy/dx)]/[(1+x^5y)^2] + x5y^4(dy/dx) = y^5-(5x^4y)/[(1+x^5y)^2]

I am stuck here. Should I be removing the denominator before removing the dy/dx and if so, that would also remove it on the other side of the equation right? I have tried to solve this problem 13 times. I am pretty frustrated.

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  1. I find it a lot more readable just to use y', instead of all those dy/dx. And, there's no reason to rearrange the equation.

    arctan(x^5y)=xy^5
    1/(1+x^10y^2) (5x^4 + x^5y') = y^5 + 5xy^4y'

    5x^4y/(1+x^10y^2) + x^5y'/(1+x^10y^2) = y^5 + 5xy^4y'

    Now collect the y' stuff

    (x^5/(1+x^10y^2) - 5xy^4)y' = y^5 - 5x^4y/(1+x^10y^2)

    Clear the fractions. You can do this without worry, because you know the denominator is never zero.

    (x^5 - 5xy^4(1+x^10y^2))y' = (y^5(1+x^10y^2) - 5x^4)

    y' = (y^5(1+x^10y^2) - 5x^4y)/(x^5 - 5xy^4(1+x^10y^2))

    and you can expand those polynomials if you want. No sweat, just a lot of algebra I.

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