What is the derivative of e^xInx ?

The derivative of e^xInx is nI(e^xx+e^x)

What is the first derivative of e^xInx ?

The first derivative of e^xInx is nI(e^xx+e^x)

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\frac{d}{d x} (e^{x} I n x)

n I (e^{x} x + e^{x})

Solution steps

\frac{d}{d x} (e^{x} I n x)

Treat

n, I

as a constant

Take the constant out:

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

= I n \frac{d}{d x} (e^{x} x)

Apply the Product Rule:

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

f = e^{x}, g = x

= I n (\frac{d}{d x} (e^{x}) x + \frac{d x}{d x} e^{x})

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

\frac{d x}{d x} = 1

= I n (e^{x} x + 1 \cdot e^{x})

Simplify

= n I (e^{x} x + e^{x})

What is the derivative of e^xInx ?
The derivative of e^xInx is nI(e^xx+e^x)
What is the first derivative of e^xInx ?
The first derivative of e^xInx is nI(e^xx+e^x)