Question

how to differentiate y^3+27=0

Answer 1

with respect to?

Answer 2

respect to y

Answer 3

is y a function of x?

Answer 4

yeah

Answer 5

so really you want dy/dx?

Answer 6

yeah

Answer 7

so use implicit differentiation

Answer 8

by how

Answer 9

its about linear equation of higher order
i need to find the roots of the equation

Answer 10

like the chain rule...
y = f(x)

you have y^3 + 27 = 0 => (f(x))^3 + 27 = 0

if you differentiate with respect to x, you first take the derivative of the outside function and then multiply by the derivative of the inside function.

Answer 11

y^3 + 27 = (y+3)(y^2-3y+9)

Answer 12

you should know how to do implicit differentiation if you're in a differential eqns class.

Answer 13

how u get that
i forgot the formula already

Answer 14

can u teach me how u get that?

Answer 15

sum of cubes, difference of cubes factoring...
http://www.purplemath.com/modules/specfact2.htm

Answer 16

Use the chain rule, the first function is \[y^{3}\], it's derivative w.r.t. x would be \[\frac{ d(y^{3}) }{ dx }\] turns out to be \[3y^{2}\] The next function is y, it's derivative w.r.t. x would be \[\frac{ dy }{ dx }\]

Note that if the function was \[x^{3}\] the first function would be \[\frac{ d(x^{3}) }{ dx } = 3x^{2}\] the next function would be of \[x\] and it's derivative is \[\frac{ dx }{ dx }=1\]

When you are differentiation a function of a variable with respect to that variable itself, it seems pretty easy because in the end you're left with \[\frac{ dx }{ dx }\] which is just one, this may confuse people as it may unknown to them that there is such a factor at the end . Note that in case of a function of a different variable,say y, the factor becomes \[\frac{ dy }{ dx }\] Also note that you cannot further solve this as you're not given y

Answer 17

However you can calculate \[\frac{ dy }{ dx }\] by rearranging the terms

Answer 18

thanks guys appreciate a lot