What is (\partial)/(\partial x)(2xe^{4xy}) ?

The answer to (\partial)/(\partial x)(2xe^{4xy}) is 2(e^{4xy}+4e^{4xy}xy)

(\partial)/(\partial x)(2xe^{4xy})

\frac{\partial}{\partial x} (2 x e^{4 x y})

2 (e^{4 x y} + 4 e^{4 x y} x y)

Solution steps

\frac{\partial}{\partial x} (2 x e^{4 x y})

Treat

y

as a constant

Take the constant out:

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

= 2 \frac{\partial}{\partial x} (x e^{4 x y})

Apply the Product Rule:

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

f = x, g = e^{4 x y}

= 2 (\frac{\partial}{\partial x} (x) e^{4 x y} + \frac{\partial}{\partial x} (e^{4 x y}) x)

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

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

= 2 (1 \cdot e^{4 x y} + e^{4 x y} \cdot 4 y x)

Simplify

= 2 (e^{4 x y} + 4 e^{4 x y} x y)

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

\int_{0}^{1} x^{2} (x + 1)^{6} d x

in v erse l a pl a ce \frac{6}{s ( s + 3 )}

\frac{d}{d x} (a sin^{3} (x))

x \to 0 - lim (\frac{14 x ^{3} + 15 x}{x})

What is (\partial)/(\partial x)(2xe^{4xy}) ?
The answer to (\partial)/(\partial x)(2xe^{4xy}) is 2(e^{4xy}+4e^{4xy}xy)