implicit differentiation check: (y/x^3)+(x/y^3)=x^2y^4

[jinh0o]

Hello, I got another problem I would like some revision on.

I believe I went the long route because I did not know I could just simplify first before taking the derivatives. So now I got to prove the long way works as well.

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sorry for making this look messy, i do not know how to properly use this forum.

The problem states:
(y/x^3)+(x/y^3)=x^2y^4

I made the (y/x^3) into (yx^-3) for the power rule and then multiplied by common denominator:


[(-3y+xy')/x^4]+[y-3xy'/y^4]=4y^3x^2y'+2y^4x


[-3y^5+xy^4y'+x^4y-3x^5y']/[x^4y^4]=4y^3x^2y'+2y^4x


multiplied both sides by [x^4y^4]:

-3y^5+xy^4y'+x^4y-3x^5y=4y^7x^6y'+2y^8x^5


get the y' to one side:

xy^4y'-3x^5y'-4y^7x^6y'=2y^8x^5+3y^5-x^4y

factor out left side by y'x:

y'x(y^4-3x^4-4y^7x^5)=2y^8x^5+3y^5-x^4y


get the y' alone:


y'=[2y^8x^5+3y^5-x^4y]/x(y^4-3x^4-4y^7x^5)


,factor out (-1), Simplify, rearrange:


y'= [y(x^4-3y^4-2y^7x^5)/x(3x^4-y^4+4y^7x^5)]




thank you

 

[stapel]

Sorry for making this look messy; I do not know how to properly use [the formatting available on] this forum.

I have typeset what I think you mean:

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The exercise states:

. . . . .\(\displaystyle \mbox{Differentiate, and solve for }\, \dfrac{dy}{dx}:\, \dfrac{y}{x^3}\, +\, \dfrac{x}{y^3}\, =\, x^2\, y^4\)

I converted the denominator of the first term to a negative power, did the same thing for the denominator of the second term:

. . . . .$\displaystyle y\, x^{-3}\, +\, x\, y^{-3}\, =\, x^2\, y^4$

...then differentiated:

. . . . .$\displaystyle \left[\, (y')\, (x^{-3})\, +\, (y)\, (-3x^{-2})\, \right]\, +\, \left[\, (1)\, (y^{-3})\, +\, (x)\, (-3y^{-2}\, y')\, \right]\, =\, (2x)\, (y^4)\, +\, (x^2)\, (4y^3\, y')$

...and then converted to postive-exponent form:

. . . . .$\displaystyle \dfrac{y'}{x^3}\, +\, \dfrac{y}{-3x^2}\, +\, \dfrac{1}{y^3}\, +\, \dfrac{x}{-3y^2}\, =\, 2xy^4\, +\, 4x^2y^3\, y'$

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However, I don't understand what happened after this point...? You say that you "then multiplied by common denominator", but this is clearly not what happened next. Please reply with clarification. Thank you!