If
x
2
25
+
y
2
64
=
1
\displaystyle {\frac{{{x}^{{2}}}}{{{25}}}}+{\frac{{{y}^{{2}}}}{{{64}}}}={1}
25
x
2
+
64
y
2
=
1
and
y
(
3
)
=
6.4
\displaystyle {y}{\left({3}\right)}={6.4}
y
(
3
)
=
6.4
, find
y
′
(
3
)
\displaystyle {y}'{\left({3}\right)}
y
′
(
3
)
by implicit differentiation.
y
′
(
3
)
\displaystyle {y}'{\left({3}\right)}
y
′
(
3
)
=
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