- #1
trancefishy
- 75
- 0
i am working on a homework assignment. it's easy, or, so i think...
Given.
[tex] 3x^2 - xy^3 + sin(x^3 - y) = 4 [/tex]
Find [tex] \frac{dy}{dx} [/tex]
not a problem. i ended up with
[tex] \frac{dy}{dx} = \frac {6x - y^3 + 3x^2 cos(x^3 - y)}{3xy^2 + cos(x^3 - y)} [/tex]
using implicit differention.
now, "What is the slope of the curve defined by the equation at the point [tex] (0, \pi) [/tex]?"
at first, i plugged it into the derivative, ended up with [tex] \pi^3 [/tex], but, upon inspection, i noticed that the point [tex] (0, \pi) [/tex] does not exist in the original equation. am I to find the slope of the curve of the derivative, or is this a typo, or am i looking at this problem incorrectly? thanks.
Given.
[tex] 3x^2 - xy^3 + sin(x^3 - y) = 4 [/tex]
Find [tex] \frac{dy}{dx} [/tex]
not a problem. i ended up with
[tex] \frac{dy}{dx} = \frac {6x - y^3 + 3x^2 cos(x^3 - y)}{3xy^2 + cos(x^3 - y)} [/tex]
using implicit differention.
now, "What is the slope of the curve defined by the equation at the point [tex] (0, \pi) [/tex]?"
at first, i plugged it into the derivative, ended up with [tex] \pi^3 [/tex], but, upon inspection, i noticed that the point [tex] (0, \pi) [/tex] does not exist in the original equation. am I to find the slope of the curve of the derivative, or is this a typo, or am i looking at this problem incorrectly? thanks.