What is the derivative of log_{10}(|1-2x^2|) ?

The derivative of log_{10}(|1-2x^2|) is -(4x)/(ln(10)(-2x^2+1))

What is the first derivative of log_{10}(|1-2x^2|) ?

The first derivative of log_{10}(|1-2x^2|) is -(4x)/(ln(10)(-2x^2+1))

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\frac{d}{d x} (lo g_{10} (1 - 2 x^{2}))

- \frac{4 x}{ln ( 10 ) ( - 2 x ^{2} + 1 )}

Solution steps

\frac{d}{d x} (lo g_{10} (1 - 2 x^{2}))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln 1 - 2 x ^{2}}{ln ( 10 )})

Take the constant out:

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

= \frac{1}{ln ( 10 )} \frac{d}{d x} (ln 1 - 2 x^{2})

Apply the chain rule:

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

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

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

= \frac{1}{ln ( 10 )} \cdot \frac{1}{∣ 1 - 2 x ^{2} ∣} (- \frac{4 x ( 1 - 2 x ^{2} )}{∣ 1 - 2 x ^{2} ∣})

Simplify

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

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

(1 + ln (x) + \frac{y}{x}) d x = (6 - ln (x)) d y

\frac{x d y}{d x} - \frac{2 d y}{d x} - (x - 2) sin (x) + y = 0

\frac{\partial}{\partial x} (4 x e^{x})

\int_{0}^{\infty} x e^{- a x} d x

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

What is the derivative of log_{10}(|1-2x^2|) ?
The derivative of log_{10}(|1-2x^2|) is -(4x)/(ln(10)(-2x^2+1))
What is the first derivative of log_{10}(|1-2x^2|) ?
The first derivative of log_{10}(|1-2x^2|) is -(4x)/(ln(10)(-2x^2+1))