What is the derivative of 4e^xx^{3x} ?

The derivative of 4e^xx^{3x} is 4(4e^xx^{3x}+3e^xx^{3x}ln(x))

What is the first derivative of 4e^xx^{3x} ?

The first derivative of 4e^xx^{3x} is 4(4e^xx^{3x}+3e^xx^{3x}ln(x))

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\frac{d}{d x} (4 e^{x} x^{3 x})

4 (4 e^{x} x^{3 x} + 3 e^{x} x^{3 x} ln (x))

Solution steps

\frac{d}{d x} (4 e^{x} x^{3 x})

Take the constant out:

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

= 4 \frac{d}{d x} (e^{x} x^{3 x})

Apply the Product Rule:

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

f = e^{x}, g = x^{3 x}

= 4 (\frac{d}{d x} (e^{x}) x^{3 x} + \frac{d}{d x} (x^{3 x}) e^{x})

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

\frac{d}{d x} (x^{3 x}) = 3 x^{3 x} (ln (x) + 1)

= 4 (e^{x} x^{3 x} + 3 x^{3 x} (ln (x) + 1) e^{x})

Expand

e^{x} x^{3 x} + 3 x^{3 x} (ln (x) + 1) e^{x} : 4 e^{x} x^{3 x} + 3 e^{x} x^{3 x} ln (x)

= 4 (4 e^{x} x^{3 x} + 3 e^{x} x^{3 x} ln (x))

\int_{1}^{10} \frac{2}{x ^{3}} d x

\frac{d}{d x} (\frac{3}{2 x + 1})

n = 1 \sum \infty ln (\frac{1}{n})

\int \frac{1}{( x + 1 ) \cdot x ^{2} + 2 x} d x

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

What is the derivative of 4e^xx^{3x} ?
The derivative of 4e^xx^{3x} is 4(4e^xx^{3x}+3e^xx^{3x}ln(x))
What is the first derivative of 4e^xx^{3x} ?
The first derivative of 4e^xx^{3x} is 4(4e^xx^{3x}+3e^xx^{3x}ln(x))