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what math are you in?

AP Calculus BC

what's "BC"

It's just a level of calculus.... I need to find the derivative of this equation...

solve for y and take the derivative?

This section involves implicit differentiation

ohhh

did you take the derivative?

No I'm still stuck on this problem!!

I don't know how to with those fractions

the derivative of \(\frac{x^2}{16}\) is \(\frac{x}{8}\)

Really? Wouldn't the 16 become a zero?

and the derivative with respect to \(x\) of \(-\frac{y^2}{9}\) is
\[-\frac{2y}{9}y'\]

no it is a constant, think
\[\frac{x^2}{16}=\frac{1}{16}x^2\]

Am I just using power rule here? I don't need to use quotient rule (that's what I was thinking...?)