Question

find dy/dx of the following :

6xy^3 - x^2 = 8 anyone?

Answer 1

Do you know how to start?
this is implicit differentiation, so, for 6xy^3 use chain rule for -x^2 power rule, and differentiation of 8 is 0.

Answer 2

for 6xy^3 use it with product rule

Answer 3

\[6xy^3 - x^2 = 8 \\ \\ 6[x(3)(y^2)(y')+y^3]-2x=0\]
\[18xy^2y'=2x \\ \\ y'=\frac{2x}{18xy^2} \\ \\y'=\frac{1}{9y^2}\]

Answer 4

how you get (y^1) + y^3? @.Sam.

Answer 5

Mistake strikes again :| , should be
\[6xy^3 - x^2 = 8 \\ \\ 6[x(3)(y^2)(y')+y^3]-2x=0\]
\[18xy^2y'=2x-6y^3 \\ \\ y'=\frac{2x-6y^3}{18xy^2} \\ \\y'=\frac{2(x-3y^3)}{18xy^2} \\ \\y'=\frac{x-3y^3}{9xy^2}\]

Answer 6

that's not y^1, that's y "Primed", means dy/dx

Answer 7

\[\Huge y'=\frac{dy}{dx} \\ \\ \Huge y^1 \neq \frac{dy}{dx}\]

Answer 8

You get (y^1) + y^3 from product rule
\[\Large \frac{d}{dx}(uv)=u\frac{dv}{dx} + v\frac{du}{dx} \]

Answer 9

need to find U and V again right? @.Sam.

Answer 10

Noneed, from
6xy^3,

"u" is the x, "v" is the y^3
so using product rule, leave the 6 out,
\[6[x(3)y^2(\frac{dy}{dx})+y^3(1)]\]

Answer 11

Everytime you differentiate "y" you will get dy/dx

Answer 12

I know already but im still confuse with the y^3(1) @.Sam.

Answer 13

That part is \[v\frac{du}{dx}\] From the product rule.
We said that "v" is the y^3 , and for du/dx means the differentiation of "u", differentiate "x" is 1,
so we get

y^3(1) or just y^3

Answer 14

thank you @.Sam. for helping Im already solved it :)

Answer 15

np :)