What is the derivative of 1/8 ln(5x^2-2) ?

The derivative of 1/8 ln(5x^2-2) is (5x)/(4(5x^2-2))

What is the first derivative of 1/8 ln(5x^2-2) ?

The first derivative of 1/8 ln(5x^2-2) is (5x)/(4(5x^2-2))

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\frac{d}{d x} (\frac{1}{8} ln (5 x^{2} - 2))

\frac{5 x}{4 ( 5 x ^{2} - 2 )}

Solution steps

\frac{d}{d x} (\frac{1}{8} ln (5 x^{2} - 2))

Take the constant out:

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

= \frac{1}{8} \frac{d}{d x} (ln (5 x^{2} - 2))

Apply the chain rule:

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

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

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

= \frac{1}{8} \cdot \frac{1}{5 x ^{2} - 2} \cdot 10 x

Simplify

\frac{1}{8} \cdot \frac{1}{5 x ^{2} - 2} \cdot 10 x : \frac{5 x}{4 ( 5 x ^{2} - 2 )}

= \frac{5 x}{4 ( 5 x ^{2} - 2 )}

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

6 \frac{d y}{d x} - 2 y = x y^{4}, y (0) = - 2

(\frac{2 y}{x} + 2 x) + (2 ln (x) - 3) (\frac{d y}{d x}) = 0

\frac{1}{4 ^{x}}

What is the derivative of 1/8 ln(5x^2-2) ?
The derivative of 1/8 ln(5x^2-2) is (5x)/(4(5x^2-2))
What is the first derivative of 1/8 ln(5x^2-2) ?
The first derivative of 1/8 ln(5x^2-2) is (5x)/(4(5x^2-2))