Question

would the derivative of (x^2/a^2 + y^2/b^2=1) be x^2+y^2=0

Answer 1

i am doing implicit differentiation

Answer 2

derivative with respect to \(x\)?

Answer 3

yes

Answer 4

start with
\[\frac{2x}{a^2}+\frac{2y}{b^2}y'=0\] then solve for \(y'\)

Answer 5

ok but my professor said we can use the constant multiple rule. how would that apply

Answer 6

i am not sure what "constant multiple" rule means, but if it means \((cf)'=cf'\) then that is why the \(a^2\) and \(b^2\) stay in the denominator since
\[\frac{x^2}{a^2}=\frac{1}{a^2}\times x^2\]

Answer 7

rule is [cf(x)]'=cf'

Answer 8

@satellite73 ?

Answer 9

what do i do after i have solved for y' but still have the a^2 and b^2

Answer 10

yes they are still there
they are just number, like 9 and 25 for example

Answer 11

subtract \(\frac{2x}{a^2}\) from both sides, then multiply both sides by \(\frac{b^2}{2y}\)

Answer 12

yes i know i got the answer. but what am i supposed to do with the constants?