What is the derivative of (6x^{4x}) ?

The derivative of (6x^{4x}) is ln(6)*81^x*2^{4x+2}x^{4x}+81^x*2^{2+4x}x^{4x}(ln(x)+1)

What is the first derivative of (6x^{4x}) ?

The first derivative of (6x^{4x}) is ln(6)*81^x*2^{4x+2}x^{4x}+81^x*2^{2+4x}x^{4x}(ln(x)+1)

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\frac{d}{d x} ((6 x)^{4 x})

ln (6) \cdot 8 1^{x} \cdot 2^{4 x + 2} x^{4 x} + 8 1^{x} \cdot 2^{2 + 4 x} x^{4 x} (ln (x) + 1)

Solution steps

\frac{d}{d x} ((6 x)^{4 x})

Simplify

(6 x)^{4 x} : 129 6^{x} x^{4 x}

= \frac{d}{d x} (129 6^{x} x^{4 x})

Apply the Product Rule:

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

f = 129 6^{x}, g = x^{4 x}

= \frac{d}{d x} (129 6^{x}) x^{4 x} + \frac{d}{d x} (x^{4 x}) 129 6^{x}

\frac{d}{d x} (129 6^{x}) = ln (6) \cdot 8 1^{x} \cdot 2^{4 x + 2}

\frac{d}{d x} (x^{4 x}) = 4 x^{4 x} (ln (x) + 1)

= ln (6) \cdot 8 1^{x} \cdot 2^{4 x + 2} x^{4 x} + 4 x^{4 x} (ln (x) + 1) \cdot 129 6^{x}

4 x^{4 x} (ln (x) + 1) \cdot 129 6^{x} = 8 1^{x} \cdot 2^{2 + 4 x} x^{4 x} (ln (x) + 1)

= ln (6) \cdot 8 1^{x} \cdot 2^{4 x + 2} x^{4 x} + 8 1^{x} \cdot 2^{2 + 4 x} x^{4 x} (ln (x) + 1)

\int \frac{1}{( 5 x - 2 ) ^{\frac{5}{2}}} d x

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

(y^{3} - 1) e^{x} d x + 3 y^{2} (e^{x} + 1) d y = 0

(2, - 7), (- 5, 2)

x \to 5 lim (x^{2} - 5)

What is the derivative of (6x^{4x}) ?
The derivative of (6x^{4x}) is ln(6)*81^x*2^{4x+2}x^{4x}+81^x*2^{2+4x}x^{4x}(ln(x)+1)
What is the first derivative of (6x^{4x}) ?
The first derivative of (6x^{4x}) is ln(6)*81^x*2^{4x+2}x^{4x}+81^x*2^{2+4x}x^{4x}(ln(x)+1)