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
e
x
⋅
y
2
+
x
5
=
sin
(
y
)
\displaystyle {e}^{{x}}\cdot{y}^{{2}}+{x}^{{5}}={\sin{{\left({y}\right)}}}
e
x
⋅
y
2
+
x
5
=
sin
(
y
)
.
d
y
d
x
=
\displaystyle \frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=\
d
x
d
y
=
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