If $x^2-xy+2y^2=\frac{a^2}{7}$, find $y'''$.

Question

If $x^2-xy+2y^2=\frac{a^2}{7}$, find $y'''$.

For our 1st derivative we got $$y'=\frac{2x-y}{x-4y}.$$

For the second derivative we got $$y''=\frac{14x^2-14xy+28y^2}{(x-4y)^3}.$$

And for the final answer we got $$y'''=\frac{4(-84x^3+119x^2y-154xy^2-84y^3)}{(x-4y)^5}.$$

Took me 2 hours and 3 white boards to get to that answer, and the teacher said another answer was right, also that there was an $a$ in the final answer, even though the derivative of a constant is zero. any insight or help is greatly appreciated, and will try to post all my work as an answer.

Answer 1

A suggestion I would make in dealing with an implicit differentiation of this sort is to not differentiate the result for $y'$ immediately, but to differentiate implicitly the equation you developed. Thus,

$$x^2 - xy + 2y^2 \ = \ \frac{a^2}{7} \ \Rightarrow \ 2x - y - xy' + 4yy' \ = \ 0 $$

[you can extract a result for $y'$ , but don't use it yet]; differentiating again:

$$\ \Rightarrow 2 -y' - y' - xy'' + 4(y')^2 + 4yy'' \ = \ 0 \ \Rightarrow \ 2 -2y' + 4(y')^2 + (4y-x) \cdot y'' \ = \ 0 \ ;$$

and once more:

$$ \Rightarrow -2y'' + 8y'y'' + (4y' - 1) \cdot y'' + (4y-x) \cdot y''' \ = \ 0 $$

$$\Rightarrow 3 \cdot (4y' - 1) \cdot y'' + (4y-x) \cdot y''' \ = \ 0 \ . $$

Replace results for derivatives as you go and then solve for $y'''$; try to avoid writing ratios as much as possible.

The Quotient Rule is no friend in a situation like this; instead, you want to save your first and second derivative results to insert later (and you'll find some simplifications by cancellation as you go, I believe).

As for how the $\frac{a^2}{7}$ gets back in there, look at your result for the second derivative:

$$y'' \ = \ \frac{14x^2 - 14xy + 28y^2}{(x - 4y)^3} \ = \ \ \frac{14 \cdot (x^2 - xy + 2y^2)}{(x - 4y)^3} \ = \ 14 \cdot \frac{\frac{a^2}{7}}{(x - 4y)^3} $$ $$= \ \frac{2a^2}{(x - 4y)^3} . $$

The "7" was not in there just to be random (although this is a somewhat messy calculation)...

Answer 2

As Will Jagy noted, $$y''=\frac{2a^2}{(x-4y)^3}.$$ Then $$y'''=2a^2(-3)(x-4y)^{-4}(1-4y')$$ Now $$1-4y'=\frac{4y-8x-4y+x}{x-4y}=\frac{7x}{x-4y},$$ so your $y'''$ is $$y'''=\frac{42a^2x}{(x-4y)^5}.$$ I used your answer for second derivative to find the third derivative and I got $$y'''=\frac{(x-4y)(28x-14y-14xy'+56yy')-(14x^2-14xy+28y^2)\times 3(1-4y')}{(x-4y)^4}$$ after cancelling $(x-4y)^2$ out from numerator and denominator. Then I substitute $y'=\frac{2x-y}{x-4y}$ and simplify, and I got $$\frac{294(x^2-xy+2y^2)x}{(x-4y)^5}$$ which agrees with my answer upon substituting $x^2-xy+2y^2=a^2/7$ . This means that if you had calculated correctly, your answer should have been $$y'''=\frac{294(x^2-xy+2y^2)x}{(x-4y)^5}.$$