What is the derivative of (2^x^8) ?

The derivative of (2^x^8) is ln(2)*2^{8x+3}

What is the first derivative of (2^x^8) ?

The first derivative of (2^x^8) is ln(2)*2^{8x+3}

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\frac{d}{d x} ((2^{x})^{8})

ln (2) \cdot 2^{8 x + 3}

Solution steps

\frac{d}{d x} ((2^{x})^{8})

Simplify

(2^{x})^{8} : 25 6^{x}

= \frac{d}{d x} (25 6^{x})

Apply the derivative exponent rule:

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

= 25 6^{x} ln (256)

Simplify

25 6^{x} ln (256) : ln (2) \cdot 2^{8 x + 3}

= ln (2) \cdot 2^{8 x + 3}

f (t) = 2 t

2 x^{5} - 3 e^{6 x}

\frac{\partial}{\partial x} (ln (3 x + 3 c t))

(x - 1) (x^{2} + 2)^{3}

y^{^{'}} + \frac{y}{2 x} = \frac{x}{y ^{3}}

What is the derivative of (2^x^8) ?
The derivative of (2^x^8) is ln(2)*2^{8x+3}
What is the first derivative of (2^x^8) ?
The first derivative of (2^x^8) is ln(2)*2^{8x+3}