What is the derivative of log_{4}(5x^2+1) ?

The derivative of log_{4}(5x^2+1) is (5x)/(ln(2)(5x^2+1))

What is the first derivative of log_{4}(5x^2+1) ?

The first derivative of log_{4}(5x^2+1) is (5x)/(ln(2)(5x^2+1))

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\frac{d}{d x} (lo g_{4} (5 x^{2} + 1))

\frac{5 x}{ln ( 2 ) ( 5 x ^{2} + 1 )}

Solution steps

\frac{d}{d x} (lo g_{4} (5 x^{2} + 1))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln ( 5 x ^{2} + 1 )}{ln ( 4 )})

Take the constant out:

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

= \frac{1}{ln ( 4 )} \frac{d}{d x} (ln (5 x^{2} + 1))

Apply the chain rule:

\frac{1}{5 x ^{2} + 1} \frac{d}{d x} (5 x^{2} + 1)

= \frac{1}{5 x ^{2} + 1} \frac{d}{d x} (5 x^{2} + 1)

\frac{d}{d x} (5 x^{2} + 1) = 10 x

= \frac{1}{ln ( 4 )} \cdot \frac{1}{5 x ^{2} + 1} \cdot 10 x

Simplify

\frac{1}{ln ( 4 )} \cdot \frac{1}{5 x ^{2} + 1} \cdot 10 x : \frac{5 x}{ln ( 2 ) ( 5 x ^{2} + 1 )}

= \frac{5 x}{ln ( 2 ) ( 5 x ^{2} + 1 )}

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

\frac{6}{r ^{3}}

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

\int \frac{ln ( n )}{n ^{2}}

\frac{4}{5 x + 1}

What is the derivative of log_{4}(5x^2+1) ?
The derivative of log_{4}(5x^2+1) is (5x)/(ln(2)(5x^2+1))
What is the first derivative of log_{4}(5x^2+1) ?
The first derivative of log_{4}(5x^2+1) is (5x)/(ln(2)(5x^2+1))