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

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

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

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

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

Solution steps

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

Simplify

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

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

n = 1 \sum \infty (sin (8))^{n}

\int x^{2} 25 - x^{2} d x

19 - x^{2}, - 2 x + 16, - 5, 0

(12 x - x^{3} + 1), x^{2} + 29, {0, 7}

x \to 0 + lim (4 + ln (x))

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