What is the derivative of log_{2}(3x^2-5) ?

The derivative of log_{2}(3x^2-5) is (6x)/(ln(2)(3x^2-5))

What is the first derivative of log_{2}(3x^2-5) ?

The first derivative of log_{2}(3x^2-5) is (6x)/(ln(2)(3x^2-5))

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derivative of

lo g_{2} (3 x^{2} - 5)

\frac{6 x}{ln ( 2 ) ( 3 x ^{2} - 5 )}

Solution steps

\frac{d}{d x} (lo g_{2} (3 x^{2} - 5))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln ( 3 x ^{2} - 5 )}{ln ( 2 )})

Take the constant out:

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

= \frac{1}{ln ( 2 )} \frac{d}{d x} (ln (3 x^{2} - 5))

Apply the chain rule:

\frac{1}{3 x ^{2} - 5} \frac{d}{d x} (3 x^{2} - 5)

= \frac{1}{3 x ^{2} - 5} \frac{d}{d x} (3 x^{2} - 5)

\frac{d}{d x} (3 x^{2} - 5) = 6 x

= \frac{1}{ln ( 2 )} \cdot \frac{1}{3 x ^{2} - 5} \cdot 6 x

Simplify

\frac{1}{ln ( 2 )} \cdot \frac{1}{3 x ^{2} - 5} \cdot 6 x : \frac{6 x}{ln ( 2 ) ( 3 x ^{2} - 5 )}

= \frac{6 x}{ln ( 2 ) ( 3 x ^{2} - 5 )}

\frac{s}{2 s + 1}

\frac{d y}{d x} + \frac{3}{x} y = 16 y^{3}, y (1) = 1

\int_{- 2}^{2} 2 - 4 - x^{2} d x

\frac{1}{s + 5}

f (x) = - 5 x^{2}, at x = 8

What is the derivative of log_{2}(3x^2-5) ?
The derivative of log_{2}(3x^2-5) is (6x)/(ln(2)(3x^2-5))
What is the first derivative of log_{2}(3x^2-5) ?
The first derivative of log_{2}(3x^2-5) is (6x)/(ln(2)(3x^2-5))