What is the derivative of ze^z ?

The derivative of ze^z is e^z+e^zz

What is the first derivative of ze^z ?

The first derivative of ze^z is e^z+e^zz

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derivative of

z e^{z}

e^{z} + e^{z} z

Solution steps

\frac{d}{d z} (z e^{z})

Apply the Product Rule:

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

f = z, g = e^{z}

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

\frac{d z}{d z} = 1

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

= 1 \cdot e^{z} + e^{z} z

Simplify

= e^{z} + e^{z} z

\frac{\partial}{\partial x} ((x^{2} - 1) \cdot (y + 2))

x \to - 8 - lim (x^{2} - 64)

n = 1 \sum \infty 4^{n + 1} 5^{- n}

\frac{d y}{d x}, y = 8 x

\frac{d}{d x} (- 2 e^{3 x})