What is (\partial)/(\partial y)(e^ysin(x)y) ?

The answer to (\partial)/(\partial y)(e^ysin(x)y) is sin(x)(e^yy+e^y)

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(\partial)/(\partial y)(e^ysin(x)y)

\frac{\partial}{\partial y} (e^{y} sin (x) y)

sin (x) (e^{y} y + e^{y})

Solution steps

\frac{\partial}{\partial y} (e^{y} sin (x) y)

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = e^{y}, g = y

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

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

\frac{\partial}{\partial y} (y) = 1

= sin (x) (e^{y} y + 1 \cdot e^{y})

Simplify

= sin (x) (e^{y} y + e^{y})

\int \frac{x ^{2} sin ( x ) - 2 x + 3}{x ^{2}} d x

\frac{d y}{d t} = - \frac{6}{( 2000 + 6 t )}

\int \frac{e ^{x}}{e ^{2 x} - 1} d x

(\frac{1}{2})^{x} - x

\int_{- 1}^{1} \frac{3}{4} x^{2} (1 - x^{2}) d x

What is (\partial)/(\partial y)(e^ysin(x)y) ?
The answer to (\partial)/(\partial y)(e^ysin(x)y) is sin(x)(e^yy+e^y)