What is (\partial)/(\partial x)(3xze^{4xy}) ?

The answer to (\partial)/(\partial x)(3xze^{4xy}) is 3z(e^{4xy}+4e^{4xy}xy)

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(\partial)/(\partial x)(3xze^{4xy})

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

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

Solution steps

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

Treat

z, y

as constants

Take the constant out:

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

= 3 z \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}

= 3 z (\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

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

Simplify

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

x \to - 3 lim (3 x^{2} - 4 x + 1)

(x + 1) (\frac{d y}{d x}) + (2 x - 1) y = e^{- 2 x}

x \to - 1 lim ((x + 2 x^{3})^{5})

f (x) = \frac{1}{2} - x^{2}, at x = - \frac{1}{2}

x \to + (- 5) lim (3 x - ∣ x + 5 ∣)

What is (\partial)/(\partial x)(3xze^{4xy}) ?
The answer to (\partial)/(\partial x)(3xze^{4xy}) is 3z(e^{4xy}+4e^{4xy}xy)