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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

= 4 \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}

= 4 (\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

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

Simplify

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

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

\int_{1}^{2} 2 (1 - \frac{1}{x ^{2}}) d x

d er i v a t i v e f (x) = \frac{x}{x ^{2} + 7}

\frac{d}{d x} (\frac{3}{4 ( 1 - x ) ^{\frac{5}{2}}})

4 x^{2} y^{^{''}} + y = 0

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