How do I find dy/dx in terms of x and y of the equation x^(5)+y^(3)+2xy^(2)=33?

Question

anonymous · Answer

jts differentiate it,answer will come itself

anonymous · Answer

$$5x^4+3y^2y'+2(y^2+2xy')=0$$ solve for $y'$

anonymous · Answer

How did you get $$2(y ^{2}+2xy')?$$

anonymous · Answer

product rule and of course the chain rule
i can explain further if you like

anonymous · Answer

oh no i can't because i made a mistake
good catch

anonymous · Answer

should be $2(y ^{2}+2xyy')$

anonymous · Answer

Okay, that's what I got, but how do you know where to put y'?

anonymous · Answer

you are thinking of $y$ as some (unspecified) function of $x$ i.e. $y=y(x)$

anonymous · Answer

but suppose for a moment that we knew $y=\sin(x)$ 
then how would you take the derivative of 
$$2x\sin^2(x)?$$
you need the product rule and the chain rule

anonymous · Answer

you would get 
$$2(\sin^2(x)+2x\sin(x)\cos(x))$$

anonymous · Answer

we don't know what $y$ is, but if i replace $\sin(x)$ by $y$ in the above, i get 
$$2(y^2+2xyy')$$

anonymous · Answer

So I solved for y'..does it look right?
$$\frac{ (-5x ^{4} + 2y ^{2} ) }{ (3y ^{2} + 4xy) }$$

anonymous · Answer

i don't know i didn't actually do it

anonymous · Answer

yeah that looks good