What is the derivative of e^{(x+ln(x)}) ?

The derivative of e^{(x+ln(x)}) is e^xx+e^x

What is the first derivative of e^{(x+ln(x)}) ?

The first derivative of e^{(x+ln(x)}) is e^xx+e^x

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

e^{x} x + e^{x}

Solution steps

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

Simplify

e^{(x + l n (x))} : e^{x} x

= \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

= \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

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

Simplify

= e^{x} x + e^{x}

y = sin (x), y = sin (2 x), x = 0, x = \frac{π}{3}

\frac{d}{d x} (1 - ln (cos (x)))

f (x) = \frac{x ^{2} - 4}{x - 3}

x \to - \infty lim (a^{x})

\int y^{2} ln (y) d y

What is the derivative of e^{(x+ln(x)}) ?
The derivative of e^{(x+ln(x)}) is e^xx+e^x
What is the first derivative of e^{(x+ln(x)}) ?
The first derivative of e^{(x+ln(x)}) is e^xx+e^x