What is the derivative of log_{4}(x^2+3) ?

The derivative of log_{4}(x^2+3) is x/(ln(2)(x^2+3))

What is the first derivative of log_{4}(x^2+3) ?

The first derivative of log_{4}(x^2+3) is x/(ln(2)(x^2+3))

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\frac{d}{d x} (lo g_{4} (x^{2} + 3))

\frac{x}{ln ( 2 ) ( x ^{2} + 3 )}

Solution steps

\frac{d}{d x} (lo g_{4} (x^{2} + 3))

Apply log rule:

lo g_{a} (b) = \frac{ln ( b )}{ln ( a )}

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the chain rule:

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

= \frac{1}{x ^{2} + 3} \frac{d}{d x} (x^{2} + 3)

\frac{d}{d x} (x^{2} + 3) = 2 x

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

Simplify

\frac{1}{ln ( 4 )} \cdot \frac{1}{x ^{2} + 3} \cdot 2 x : \frac{x}{ln ( 2 ) ( x ^{2} + 3 )}

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

\int \frac{1}{x ( x + 1 )} d x

x^{3} - 3 x^{2} + 2 x - 2, at x = 2

\frac{d}{d x} (2 - e^{x})

\frac{d y}{d x} - \frac{2}{x} y = x

\int (e^{3.1})^{2} d x

What is the derivative of log_{4}(x^2+3) ?
The derivative of log_{4}(x^2+3) is x/(ln(2)(x^2+3))
What is the first derivative of log_{4}(x^2+3) ?
The first derivative of log_{4}(x^2+3) is x/(ln(2)(x^2+3))