What is the derivative of (6x^{3x}) ?

The derivative of (6x^{3x}) is ln(6)*8^x*3^{3x+1}x^{3x}+8^x*3^{1+3x}x^{3x}(ln(x)+1)

What is the first derivative of (6x^{3x}) ?

The first derivative of (6x^{3x}) is ln(6)*8^x*3^{3x+1}x^{3x}+8^x*3^{1+3x}x^{3x}(ln(x)+1)

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\frac{d}{d x} ((6 x)^{3 x})

ln (6) \cdot 8^{x} \cdot 3^{3 x + 1} x^{3 x} + 8^{x} \cdot 3^{1 + 3 x} x^{3 x} (ln (x) + 1)

Solution steps

\frac{d}{d x} ((6 x)^{3 x})

Simplify

(6 x)^{3 x} : 21 6^{x} x^{3 x}

= \frac{d}{d x} (21 6^{x} x^{3 x})

Apply the Product Rule:

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

f = 21 6^{x}, g = x^{3 x}

= \frac{d}{d x} (21 6^{x}) x^{3 x} + \frac{d}{d x} (x^{3 x}) 21 6^{x}

\frac{d}{d x} (21 6^{x}) = ln (6) \cdot 8^{x} \cdot 3^{3 x + 1}

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

= ln (6) \cdot 8^{x} \cdot 3^{3 x + 1} x^{3 x} + 3 x^{3 x} (ln (x) + 1) \cdot 21 6^{x}

3 x^{3 x} (ln (x) + 1) \cdot 21 6^{x} = 8^{x} \cdot 3^{1 + 3 x} x^{3 x} (ln (x) + 1)

= ln (6) \cdot 8^{x} \cdot 3^{3 x + 1} x^{3 x} + 8^{x} \cdot 3^{1 + 3 x} x^{3 x} (ln (x) + 1)

\int sec (x) tan (x) d x

y^{^{''}} + 5 y^{^{'}} = (4 t - 2) sin (2 t)

\frac{d}{d t} (3 t e^{t})

2 y^{^{''}} + 5 y^{^{'}} - 3 y = 0, y (0) = - 1, y^{^{'}} (0) = 24

\int_{- 3}^{- 1} (x + 2)^{99} d x

What is the derivative of (6x^{3x}) ?
The derivative of (6x^{3x}) is ln(6)*8^x*3^{3x+1}x^{3x}+8^x*3^{1+3x}x^{3x}(ln(x)+1)
What is the first derivative of (6x^{3x}) ?
The first derivative of (6x^{3x}) is ln(6)*8^x*3^{3x+1}x^{3x}+8^x*3^{1+3x}x^{3x}(ln(x)+1)