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

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

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

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

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derivative of

y = (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}

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

4 y^{^{''}} + 4 y^{^{'}} + 8 y = 0

x \to 0 lim (cos (9 x))

n = 1 \sum \infty n 2^{- n}

\int x y e^{y^{2} x} d y

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