What is the derivative of 8(2ze^z+e^zz^2) ?

The derivative of 8(2ze^z+e^zz^2) is 8(e^zz^2+4e^zz+2e^z)

What is the first derivative of 8(2ze^z+e^zz^2) ?

The first derivative of 8(2ze^z+e^zz^2) is 8(e^zz^2+4e^zz+2e^z)

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derivative of

8 (2 z e^{z} + e^{z} z^{2})

8 (e^{z} z^{2} + 4 e^{z} z + 2 e^{z})

Solution steps

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

Take the constant out:

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

= 8 \frac{d}{d z} (2 z e^{z} + e^{z} z^{2})

Apply the Sum/Difference Rule:

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

= 8 (\frac{d}{d z} (2 z e^{z}) + \frac{d}{d z} (e^{z} z^{2}))

\frac{d}{d z} (2 z e^{z}) = 2 (e^{z} + e^{z} z)

\frac{d}{d z} (e^{z} z^{2}) = e^{z} z^{2} + 2 z e^{z}

= 8 (2 (e^{z} + e^{z} z) + e^{z} z^{2} + 2 z e^{z})

Expand

2 (e^{z} + e^{z} z) + e^{z} z^{2} + 2 z e^{z} : e^{z} z^{2} + 4 e^{z} z + 2 e^{z}

= 8 (e^{z} z^{2} + 4 e^{z} z + 2 e^{z})

\frac{d y}{d x} = - 4 x y^{2}

x \to - 3 lim (x + 1)

x + 2, \frac{1}{x + 1}, x = 2

x \to 0 lim (\frac{e ^{x}}{x})

\frac{\partial}{\partial x} (x^{5} y z^{2} + 2 yz)

What is the derivative of 8(2ze^z+e^zz^2) ?
The derivative of 8(2ze^z+e^zz^2) is 8(e^zz^2+4e^zz+2e^z)
What is the first derivative of 8(2ze^z+e^zz^2) ?
The first derivative of 8(2ze^z+e^zz^2) is 8(e^zz^2+4e^zz+2e^z)