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

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

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

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

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

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

Solution steps

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

Apply log rule:

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

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

(2 x + \frac{2}{y}) + (\frac{( 2 x ^{2} )}{y} + \frac{2 x}{y ^{2}}) y^{^{'}} = 0

f (x) = x^{4} - 10 x^{2} + 9

\int - \frac{14}{x ^{15}} d x

\int x^{n} d x

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

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