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

The derivative of 4^{3x} is 3ln(2)*2^{6x+1}

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

The first derivative of 4^{3x} is 3ln(2)*2^{6x+1}

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

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

Solution steps

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

Simplify

4^{3 x} : 6 4^{x}

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

Apply the derivative exponent rule:

\frac{d}{d x} (a^{x}) = a^{x} ln (a)

= 6 4^{x} ln (64)

Simplify

6 4^{x} ln (64) : 3 ln (2) \cdot 2^{6 x + 1}

= 3 ln (2) \cdot 2^{6 x + 1}

\int \frac{6 x + 4}{x ^{2} ( x ^{2} + 2 )} d x

(x - 2 y) d y + (y + 4 x) d x = 0

3 x^{4} 25 - x^{2}

\int 4 sin (t) cos (t) d t

x \to 3 lim (\frac{x ^{3} - 27}{x - 3})

What is the derivative of 4^{3x} ?
The derivative of 4^{3x} is 3ln(2)*2^{6x+1}
What is the first derivative of 4^{3x} ?
The first derivative of 4^{3x} is 3ln(2)*2^{6x+1}