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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

\frac{\partial}{\partial x} (\frac{x}{2} + y) = \frac{1}{2}

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

Simplify

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

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

\int + x^{2} e^{4 x} d x

y^{^{''}} + 16 y = e^{- t}

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

9 y ln (x) - x y^{^{'}} = 0

36 y^{^{''}} + 84 y^{^{'}} + 49 y = 0

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