What is the derivative of f(x)=log_{2}(x^2+1)-2^3 ?

The derivative of f(x)=log_{2}(x^2+1)-2^3 is (2x)/(ln(2)(x^2+1))

What is the first derivative of f(x)=log_{2}(x^2+1)-2^3 ?

The first derivative of f(x)=log_{2}(x^2+1)-2^3 is (2x)/(ln(2)(x^2+1))

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derivative of

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

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

Solution steps

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

Simplify

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)

Apply the Sum/Difference Rule:

(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

Simplify

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

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

\int p (p + 6)^{8} d p

\frac{d}{d x} (\frac{0.2 x}{x ^{2} + 9})

\int 1 - x^{9} d x

\frac{\partial}{\partial x} (- \frac{1}{x})

What is the derivative of f(x)=log_{2}(x^2+1)-2^3 ?
The derivative of f(x)=log_{2}(x^2+1)-2^3 is (2x)/(ln(2)(x^2+1))
What is the first derivative of f(x)=log_{2}(x^2+1)-2^3 ?
The first derivative of f(x)=log_{2}(x^2+1)-2^3 is (2x)/(ln(2)(x^2+1))