Please check my work (extrema, implicit diff., etc)

[shivers20]

1) Find the extreme point and inflection point.
y=6-2x-x^2

Find 1st derivative: y' = -2 -2x
Equal y' to zero: -2(1-x) = (x=-1) critical point is x=-1 and x=0
Find 2nd derivative: y" = -2
Equal it to zero: -2=0 (x=0) no inflection point (0,0)
Find absolute extreme: I am stumped. Do I need to plug in -1 into original equation? (-1, 9) ?


2) Use implicit differentiation to find dy/dx
x^3 + y^3 = 18xy

Find 1st derivative: 3x^2 + 3y^2 dy/dx = 18y+18x dy/dx

. . . . . . . . . . . . . . . .3x^2-18y = -3y^2 dy/dx + 18x dy/dx

Factor:. . . . . . . . . .3(x^2-6y) = 3(-y^2+6x) dy/dx

Answer:. . . . . . . . . . . .. . . . . .= 3(x^2-6y)/ 3(-y^2=6x)dy/dx


Question: How do I know which x or y variables to choose from when picking a dy/dx. In other words how did I know to pick y^3 dy/dx instead of picking 18y dy/dx?


3) Find the derivatives:
y = (x^2+x)(x^2-x+1) / x^4

Okay, I know how to use the quotient rule, what I do no know is how to solve the top part of the equation. Do I need to do product rule for the numerator before anything else or should I factor it or should I apply the quotient rule just like this:

(x^4) (2x+1)(2x-1) - (4x^3) (x^2+x)(x^2-x+1)/(x^4)^2

Now factor & simplify: I got some crazy number =) not even worth posting.

 

[galactus]

Re: Please check my work

Find the derivatives:
y = (x^2+x)(x^2-x+1) / x^4

Okay, I know how to use the quotient rule, what I do no know is how to solve the top part of the equation. Do I need to do product rule for the numerator before anything else or should I factor it or should I apply the quotient rule just like this:

(x^4) (2x+1)(2x-1) - (4x^3) (x^2+x)(x^2-x+1)/(x^4)^2


Now factor & simplify: I got some crazy number =) not even worth posting.

Simplify the numerator:

\(\displaystyle \L\\(x^{2}+x)(x^{2}-x+1)=x^{4}-x^{3}+x^{2}+x^{3}-x^{2}+x=x^{4}+x\)

Now, you have:

\(\displaystyle \L\\\frac{x^{4}+x}{x^{4}}=1+x^{-3}\)

Ain't that somethin'?. Look what that monstrosity whittled down to?.

\(\displaystyle \text{Easy as {\pi}}\)

 

[daon]

Equal y' to zero: -2(1-x) = (x=-1) critical point is x=-1 and x=0

X=0 is not a critical point.

Factor: 3(x^2-6y) = 3(-y^2+6x) dy/dx

Answer: = 3(x^2-6y)/ 3(-y^2=6x)dy/dx

Huh?

I think you meant y' = (x^2 - 6y)/(-y^2+6x)

Question: How do I know which x or y variables to choose from when picking a dy/dx. In other words how did I know to pick y^3 dy/dx instead of picking 18y dy/dx?

Anything that has a y in it gets a dy/dx multipled on the end (assuming y is the function and x is the independent variable). This is, in all its glory, the chain rule. Since y is a function of x, when taking the derivative of a function containing y you must also take the derivative of y with respect to x which is dy/dx. In the case of the 18y you were not differentiating with respect to y, you were using the product rule.

For your derivative question, multiply out the top to get a polynomial then divide each term by x^4. You will have another simple polynomial to use the power rule with.