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

The derivative of 3^{4x} is 4ln(3)*81^x

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

The first derivative of 3^{4x} is 4ln(3)*81^x

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

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

Solution steps

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

Simplify

3^{4 x} : 8 1^{x}

= \frac{d}{d x} (8 1^{x})

Apply the derivative exponent rule:

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

= 8 1^{x} ln (81)

Simplify

ln (81) : 4 ln (3)

= 4 ln (3) \cdot 8 1^{x}

f (x) = x^{2} - 3, (- 4, 13)

\int \frac{x ^{2} - 1}{x ^{3} + x} d x

\int_{0}^{1} x e^{3 x^{2}} d x

tan (t) y^{^{'}} - y = 1, y (\frac{π}{6}) = 3

\frac{x}{x ^{2} - x + 1}

What is the derivative of 3^{4x} ?
The derivative of 3^{4x} is 4ln(3)*81^x
What is the first derivative of 3^{4x} ?
The first derivative of 3^{4x} is 4ln(3)*81^x