I want to implicitly differentiate this twice with respect to x (y is the dependent variable), (finding y''): $$(xy)^3 = 4x$$ using the derivative rules we would get $$y'' = -\frac{3x^3y^3+8x - 3x^2y^4 + 12y}{3x^4y^3}$$
but, what if we assume $$xy = n$$
the equation would be $$n^3 = 4x$$
thus $$x = \frac{n^3}{4}$$
and since $$xy = n$$, $$y = \frac{4}{n^2}$$ right?
and after we found a direct relationship between x and n on a hand, and y and n on an other hand, can't we use the chain rule like this to find the second derivative? :
$$\frac{d}{dx} = \frac{dy}{dn} \frac{dn}{dx}$$
from the relationships i made, $$\frac{dn}{dx} = \frac{4}{3n^2}$$
and $$\frac{dy}{dn} = \frac{-8}{n^3}$$
so dy/dx = $$\frac{-32}{3n^5}$$ , and if we take the second derivative using the same way it would be $$\frac{-128}{9n^7}$$
is the second method correct? (substitution by n along with chain rule)
- somebody make the expressions size bigger here please, i don't know how
I get something different for the second differentiation:
$$y'' = \frac {d}{dx} (-\frac {32}{3} n^{-5}) = \frac {160}{3}n^{-6}\frac {dn}{dx} = \frac {640}{9n^8}$$
I think more straight forward would be:
$$x^3y^3 = 4x\\ y = 4^{\frac 13} x^{\frac {-2}3}\\ y' = -4^{\frac 13}\frac {2}{3} x^{\frac {-5}3}\\ y'' = 4^{\frac 13}\frac {10}{9} x^{\frac {-8}3}\\ y'' = \frac {10y}{9x^2}\\ $$
Does $\frac {640}{9n^8} = \frac {10y}{9x^2}$?
$$\frac {640n}{9n^9}\\ \frac {640xy}{9(xy)^9}\\ \frac {640xy}{9(4x)^3}\\ \frac {10y}{9x^2}$$
and what about $-\frac{3x^3y^3+8x - 3x^2y^4 + 12y}{3x^4y^3}$?
I haven't checked your work on the differentiation.
Applying $x^3y^3 = 4x$ where able I get $-\frac{12x + 8x - 12y + 12y}{12x^2}= \frac {10}{6} x$
It looks like something happened in the differentiation
Update
$$y''=-\frac{-18x^4y^6+24x^2y^3+32}{9x^6y^5}\\ x^2y^3 = 4 -\frac{-18(16)^6+24(4)+32}{9(4x)x^3y^2}\\ \frac{160}{9(4x)x^3y^2}\\ \frac{160y}{9(4x)x^3y^3}\\ \frac{160y}{9(4x)^2}\\ \frac {10y}{x^2}$$