derivative f(x)=log_{2}(x^2+1)-2^3

Lösung

ableitung von

f (x) = lo g_{2} (x^{2} + 1) - 2^{3}

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

Schritte zur Lösung

\frac{d}{d x} (lo g_{2} (x^{2} + 1) - 2^{3})

Vereinfache

lo g_{2} (x^{2} + 1) - 2^{3} : lo g_{2} (x^{2} + 1) - 8

= \frac{d}{d x} (lo g_{2} (x^{2} + 1) - 8)

Wende die Summen-/Differenzregel an:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (lo g_{2} (x^{2} + 1)) - \frac{d}{d x} (8)

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

\frac{d}{d x} (8) = 0

= \frac{2 x}{ln ( 2 ) ( x ^{2} + 1 )} - 0

Vereinfache

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

\frac{\partial}{\partial x} (y + 3)

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