What is the derivative of 2^{4x} ?

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

What is the first derivative of 2^{4x} ?

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

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derivative of

2^{4 x}

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

Solution steps

\frac{d}{d x} (2^{4 x})

Simplify

2^{4 x} : 1 6^{x}

= \frac{d}{d x} (1 6^{x})

Apply the derivative exponent rule:

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

= 1 6^{x} ln (16)

Simplify

1 6^{x} ln (16) : ln (2) \cdot 2^{4 x + 2}

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

\int 2 x 4 x^{2} + 1 d x

n = 0 \sum \infty \frac{2}{n ^{2} + 4 n + 3}

\int x - x^{2} d x

\frac{d}{d x} (\frac{- 4 x ^{6} + 4 x ^{5} - 3 x ^{3}}{3 x ^{2}})

\int e^{2 s i n^{2} (x)} sin (2 x) d x

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