Use implicit differentiation to determine
d
y
d
x
\displaystyle \frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}
d
x
d
y
given the equation
x
6
+
y
2
=
−
1
\displaystyle {x}^{{6}}+{y}^{{2}}=-{1}
x
6
+
y
2
=
−
1
.
d
y
d
x
=
\displaystyle \frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=\
d
x
d
y
=
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