find the implicit derivative of -2x^3+(x^2)y-3y^3=-1

Question

amistre64 · Answer

its the same rules as for explicit

amistre64 · Answer

"but not when it has a product of y" is a misstatement.  the product rule still applies to terms that are a product of y

anonymous · Answer

ok sure, thanks amistre64, I know how to solve it but may have difficulty explaining it, so I will leave it to you

amistre64 · Answer

$$D[-2x^3+x^2y-3y^3=-1]$$
$$D[-2x^3]+D[x^2y]-D[3y^3]=D[-1]$$
exponent rule, product rule, exponent rule(and chain rule for the y), constant rule

anonymous · Answer

would the final answer be $$6x ^{2}/(2x-9y ^{2})$$

amistre64 · Answer

hmm, quite possible.  i like the 6x^2 aprt; and the division looks good to

amistre64 · Answer

but let me check
$$D[-2x^3]+D[x^2y]-D[3y^3]=D[-1]$$
$$-6x^2+D[x^2]y]+x^2D[y]-9y^2y'=0$$
$$-6x^2+2xy+x^2y'-9y^2y'=0$$
a little off

amistre64 · Answer

that chain rule seems to have been the culprit in yours
$$-6x^2+2xy+x^2y'-9y^2y'=0$$
$$x^2y'-9y^2y'={6x^2-2xy}$$
$$y'(x^2-9y^2)={6x^2-2xy}$$
$$y'=\frac{6x^2-2xy}{x^2-9y^2}$$