What is the derivative of y=ln((x^4-10)/x) ?

The derivative of y=ln((x^4-10)/x) is (3x^4+10)/(x(x^4-10))

What is the first derivative of y=ln((x^4-10)/x) ?

The first derivative of y=ln((x^4-10)/x) is (3x^4+10)/(x(x^4-10))

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derivative of

y = ln (\frac{x ^{4} - 10}{x})

\frac{3 x ^{4} + 10}{x ( x ^{4} - 10 )}

Solution steps

\frac{d}{d x} (ln (\frac{x ^{4} - 10}{x}))

Apply the chain rule:

\frac{1}{\frac{x ^{4} - 10}{x}} \frac{d}{d x} (\frac{x ^{4} - 10}{x})

= \frac{1}{\frac{x ^{4} - 10}{x}} \frac{d}{d x} (\frac{x ^{4} - 10}{x})

\frac{d}{d x} (\frac{x ^{4} - 10}{x}) = \frac{3 x ^{4} + 10}{x ^{2}}

= \frac{1}{\frac{x ^{4} - 10}{x}} \cdot \frac{3 x ^{4} + 10}{x ^{2}}

Simplify

\frac{1}{\frac{x ^{4} - 10}{x}} \cdot \frac{3 x ^{4} + 10}{x ^{2}} : \frac{3 x ^{4} + 10}{x ( x ^{4} - 10 )}

= \frac{3 x ^{4} + 10}{x ( x ^{4} - 10 )}

x^{3} - 3

\int 5 x^{5} d x

(1 + cos^{2} (x))^{5}

n = 0 \sum \infty \frac{( e ^{n} )}{5}

\frac{d}{d x} (- 8 x^{3})

What is the derivative of y=ln((x^4-10)/x) ?
The derivative of y=ln((x^4-10)/x) is (3x^4+10)/(x(x^4-10))
What is the first derivative of y=ln((x^4-10)/x) ?
The first derivative of y=ln((x^4-10)/x) is (3x^4+10)/(x(x^4-10))