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

1. Loser66

where are you stuck?

2. elleblythe

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

3. Loser66

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

4. Loser66

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

5. elleblythe

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

6. Loser66

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

7. Loser66

now, combine and isolate dy/dx, please

8. elleblythe

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

9. Loser66

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