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

The answer to (\partial)/(\partial y)(2e^{x/2+y}-y) is 2e^{x/2+y}-1

Popular >

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

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

2 e^{\frac{x}{2} + y} - 1

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 2 e^{\frac{x}{2} + y} - 1

\frac{490 t}{( t ^{2} + 7 ) ^{3}}

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

sin (t - 8)

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

\frac{d}{d x} (6 x^{2} sin (x) tan (x))

What is (\partial)/(\partial y)(2e^{x/2+y}-y) ?
The answer to (\partial)/(\partial y)(2e^{x/2+y}-y) is 2e^{x/2+y}-1