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

The answer to (\partial)/(\partial y)(2^x+2^y) is ln(2)*2^y

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(\partial)/(\partial y)(2^x+2^y)

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

ln (2) \cdot 2^{y}

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial y} (2^{x}) = 0

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

= 0 + 2^{y} ln (2)

Simplify

= ln (2) \cdot 2^{y}

x \to 2 lim (\frac{x - 3}{x ^{2} - 9})

\frac{d}{d x} (\frac{e ^{x}}{e ^{x} + e ^{y} + e ^{z}})

\frac{d y}{d x} = 3 x^{7} y - y, y (1) = - 9

y^{^{''}} + 4 y^{^{'}} + 4 y = 3 e^{- 2 x}

\int tan (a x) d x

What is (\partial)/(\partial y)(2^x+2^y) ?
The answer to (\partial)/(\partial y)(2^x+2^y) is ln(2)*2^y