What is (\partial)/(\partial y)(e^ycos(x)-y^2) ?

The answer to (\partial)/(\partial y)(e^ycos(x)-y^2) is cos(x)e^y-2y

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(\partial)/(\partial y)(e^ycos(x)-y^2)

\frac{\partial}{\partial y} (e^{y} cos (x) - y^{2})

cos (x) e^{y} - 2 y

Solution steps

\frac{\partial}{\partial y} (e^{y} cos (x) - y^{2})

Treat

x

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial y} (e^{y} cos (x)) - \frac{\partial}{\partial y} (y^{2})

\frac{\partial}{\partial y} (e^{y} cos (x)) = cos (x) e^{y}

\frac{\partial}{\partial y} (y^{2}) = 2 y

= cos (x) e^{y} - 2 y

\int cos^{8} (θ) sin (θ) d θ

y - 3 = 5 (x - 2) y - 3 = 5 (x - 2)

\frac{\partial}{\partial x} (z^{2} y^{2} - e^{x})

\int \frac{26}{( 169 x ^{2} + 1 ) ^{2}} d x

\frac{\partial}{\partial y} (\frac{y}{x ^{2} - y ^{2}})

What is (\partial)/(\partial y)(e^ycos(x)-y^2) ?
The answer to (\partial)/(\partial y)(e^ycos(x)-y^2) is cos(x)e^y-2y