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

The derivative of 4ln((x/4 ^2-1)) is (8x)/(x^2-16)

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

The first derivative of 4ln((x/4 ^2-1)) is (8x)/(x^2-16)

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

\frac{8 x}{x ^{2} - 16}

Solution steps

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

Simplify

4 ln ((\frac{x}{4})^{2} - 1) : 4 ln (\frac{x ^{2}}{16} - 1)

= \frac{d}{d x} (4 ln (\frac{x ^{2}}{16} - 1))

Take the constant out:

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

= 4 \frac{d}{d x} (ln (\frac{x ^{2}}{16} - 1))

Apply the chain rule:

\frac{1}{\frac{x ^{2}}{16} - 1} \frac{d}{d x} (\frac{x ^{2}}{16} - 1)

= \frac{1}{\frac{x ^{2}}{16} - 1} \frac{d}{d x} (\frac{x ^{2}}{16} - 1)

\frac{d}{d x} (\frac{x ^{2}}{16} - 1) = \frac{x}{8}

= 4 \cdot \frac{1}{\frac{x ^{2}}{16} - 1} \cdot \frac{x}{8}

Simplify

4 \cdot \frac{1}{\frac{x ^{2}}{16} - 1} \cdot \frac{x}{8} : \frac{8 x}{x ^{2} - 16}

= \frac{8 x}{x ^{2} - 16}

x \to 0 + lim (2 + x^{9} ln (x))

\frac{\partial}{\partial y} (x y e^{yz})

\int 9 e^{u} d u

f (x) = x^{2} + x, (7, 56)

\frac{\partial}{\partial x} (\frac{7 x - 4 y}{7 x + 4 y})

What is the derivative of 4ln((x/4 ^2-1)) ?
The derivative of 4ln((x/4 ^2-1)) is (8x)/(x^2-16)
What is the first derivative of 4ln((x/4 ^2-1)) ?
The first derivative of 4ln((x/4 ^2-1)) is (8x)/(x^2-16)