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Calculus/derivatives

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Expert: - 5/7/2009


Question
QUESTION: hey, another questions on derivatives,
find dy/dx given that y^5+y^3+y-x=0

Im just not sure how to approach this with the multiple y's

ANSWER: Take the derivative with the chain rule for each term.

This means that every term with a y will be multiplied by y'.

The derivative of y^n is the same  as the derivative of x^n,
it just has a multiple of y'

The derivative is (5y^4 + 3y^2 + 1)y' - 1 = 0.

Add 1 to both sides and divide by (5y^4 + 3y^2 + 1).

That will then say y'= ....


---------- FOLLOW-UP ----------

QUESTION: does dy/dx mean that y will be in the solution or x?

just a bit stuck understanding the procedure, what part is to be divided by (5y^4+3y^2+1)? and is the formula just to be rearranged then to get the derivative of y?

Answer
When dy/dx is said, it means the derivative of y with repect to x.
It is written that way for a reason.

If x is a function of t, then the derivative of x with respect to t would be dx/dt.

That would make the derivative of y with respect to t treat these just like fraction.

dy/dt would be (dy/dx)(dx/dt).  In fractions, the dx's cancel,
so all you are left with is dy/dt.


See, if we take the equation (5y^4 + 3y^2 + 1)y' - 1 = 0
and add 1 to both sides, we get (5y^4 + 3y^2 + 1)y' = 1.
We then divide both sides by (5y^4 + 3y^2 + 1) and get
y' = 1/(5y^4 + 3y^2 + 1).

That looks like the start of diff eqs (differential equations) to me.

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