What is the derivative of ln(sqrt(x^4-4x)) ?

The derivative of ln(sqrt(x^4-4x)) is (2(x^3-1))/(x^4-4x)

What is the first derivative of ln(sqrt(x^4-4x)) ?

The first derivative of ln(sqrt(x^4-4x)) is (2(x^3-1))/(x^4-4x)

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

\frac{2 ( x ^{3} - 1 )}{x ^{4} - 4 x}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (x^{4} - 4 x) = \frac{2 ( x ^{3} - 1 )}{x ^{4} - 4 x}

= \frac{1}{x ^{4} - 4 x} \cdot \frac{2 ( x ^{3} - 1 )}{x ^{4} - 4 x}

Simplify

\frac{1}{x ^{4} - 4 x} \cdot \frac{2 ( x ^{3} - 1 )}{x ^{4} - 4 x} : \frac{2 ( x ^{3} - 1 )}{x ^{4} - 4 x}

= \frac{2 ( x ^{3} - 1 )}{x ^{4} - 4 x}

\int_{0}^{2} π (x + 1)^{2} d x

\frac{d}{d x} (x^{4} + 3 x)

\int \frac{lo g _{3} ( x )}{x} d x

y^{^{'}} + (\frac{1 + x}{x}) y = 0

y = ln (x^{2}), (5, ln (25))

What is the derivative of ln(sqrt(x^4-4x)) ?
The derivative of ln(sqrt(x^4-4x)) is (2(x^3-1))/(x^4-4x)
What is the first derivative of ln(sqrt(x^4-4x)) ?
The first derivative of ln(sqrt(x^4-4x)) is (2(x^3-1))/(x^4-4x)