What is (\partial)/(\partial x)((x+y)e^{3x-4y}) ?

The answer to (\partial)/(\partial x)((x+y)e^{3x-4y}) is e^{3x-4y}+3e^{3x-4y}(x+y)

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(\partial)/(\partial x)((x+y)e^{3x-4y})

\frac{\partial}{\partial x} ((x + y) e^{3 x - 4 y})

e^{3 x - 4 y} + 3 e^{3 x - 4 y} (x + y)

Solution steps

\frac{\partial}{\partial x} ((x + y) e^{3 x - 4 y})

Treat

y

as a constant

Apply the Product Rule:

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

f = x + y, g = e^{3 x - 4 y}

= \frac{\partial}{\partial x} (x + y) e^{3 x - 4 y} + \frac{\partial}{\partial x} (e^{3 x - 4 y}) (x + y)

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

\frac{\partial}{\partial x} (e^{3 x - 4 y}) = e^{3 x - 4 y} \cdot 3

= 1 \cdot e^{3 x - 4 y} + e^{3 x - 4 y} \cdot 3 (x + y)

Simplify

= e^{3 x - 4 y} + 3 e^{3 x - 4 y} (x + y)

\frac{d ^{2}}{d x ^{2}} (\frac{e ^{x}}{8 x + 1})

ln (ln (x^{2}))

y^{^{'}} - \frac{5}{x} y = \frac{y ^{5}}{x ^{11}}

n = 1 \sum \infty \frac{4}{1 0 ^{n}}

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

What is (\partial)/(\partial x)((x+y)e^{3x-4y}) ?
The answer to (\partial)/(\partial x)((x+y)e^{3x-4y}) is e^{3x-4y}+3e^{3x-4y}(x+y)