derivative (8x)^{2x}

Lösung

ableitung von

(8 x)^{2 x}

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

Schritte zur Lösung

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

Vereinfache

(8 x)^{2 x} : 6 4^{x} x^{2 x}

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

Wende die Produktregel an:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = 6 4^{x}, g = x^{2 x}

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

\frac{d}{d x} (6 4^{x}) = 3 ln (2) \cdot 2^{6 x + 1}

\frac{d}{d x} (x^{2 x}) = 2 x^{2 x} (ln (x) + 1)

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

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

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

\frac{\partial}{\partial s} (\frac{s - 2 r}{t})

\int \frac{3 x + 1}{x} d x

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 3, y^{^{'}} (0) = 0

\frac{d}{d x} (2 x^{2} + 7)

t an g e n t f (x) = (4 x + 5) (5 x - 1), at x = 3