I'm finding it difficult to find the second derivative of the following equation: $2x^3 + y^3 = 5$.
My answer is $\dfrac{-4xy^3 – 8x^4}{y^5}$, but the answer key says $\dfrac{-20x}{y^5}$.
My friends and I have the same answer, but I don't know where I went wrong.
Here is my work:
I got the first derivative as $\dfrac{-2x^2}{y^2}$.
I then used the quotient rule to get
$-2 \left(\dfrac{2xy^2 – 2x^2y\frac{dy}{dx}}{y^4}\right)$
After factoring out $2y$ from the numerator, I got $-4 \left(\dfrac{xy – x^2\left(\frac{-2x}{y^2}\right)}{y^3}\right)$.
After splitting the fraction into $\dfrac{x}{y^2} + \dfrac{2x^4}{y^5}$, I ended up getting
$\dfrac{-4x}{y^2}-\dfrac{-8x^4}{y^5}$.
Could somebody point out my mistake? This isn't homework by the way. I'm preparing for a placement test tomorrow, but I'm not really sure where I went wrong.
Best Answer
Your answer only requires some re-arrangement: $$ 2x^3 + y^3 = 5 $$ multiply by $x$ on both sides to get: $$ 2x^4 + xy^3 = 5 x $$ or: $$ xy^3 = 5x - 2x^4 $$ ...(substitute this in your equation to get): $$ \frac{-4(5x - 2x^4 )- 8x^4}{y^5}\to y''=\frac{-20x}{y^5} $$