Find the dy/dx of: 6x^(4/3) + 2y^5 = x^3y^2 + cube root of pi^5

Question

loser66 · Answer

where are you stuck?

elleblythe · Answer

@Loser66 is the correct answer (not simplified): -cube root of 8 / (-10y^4 + 2x^3y + 3x^2y) ?

loser66 · Answer

You guessed? or just pick one of the options and check it from me?

loser66 · Answer

Show me your work, please. 
Just take derivative both sides and isolate y'. Done.

elleblythe · Answer

@Loser66 I solved for it. $$8x ^{1/3}+10y ^{4}(dy/dx)=x^32y(dy/dx)+3x^2y^2(dy/dx)+5/3\pi^^{2/3}$$

loser66 · Answer

the left hand side is ok, but the right one!!
first term of the right one is $x^3y^2$, hence its derivative is $3x^2y^2+2x^3y (dy/dx)$
the last term is a constant, hence its derivative =0 , ignore it

loser66 · Answer

now, combine and isolate dy/dx, please

elleblythe · Answer

@Loser66  -cube root of 8 + 3x^2y^2 / (-10y^4 + 2x^3y)

loser66 · Answer

the sign of denominator is not correct, 
it should be $\dfrac{dy}{dx}=\dfrac{3x^2y^2-\sqrt[3]{x}}{10y^4-2x^3y}$