What is expand (x^4)(-(ln(x))/(4x^3)-1/(12x^3)) ?

The solution to expand (x^4)(-(ln(x))/(4x^3)-1/(12x^3)) is -(xln(x))/4-x/(12)

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expand (x^4)(-(ln(x))/(4x^3)-1/(12x^3))

expand

(x^{4}) (- \frac{ln ( x )}{4 x ^{3}} - \frac{1}{12 x ^{3}})

- \frac{x ln ( x )}{4} - \frac{x}{12}

Solution steps

(x^{4}) (- \frac{ln ( x )}{4 x ^{3}} - \frac{1}{12 x ^{3}})

Apply the distributive law:

a (b - c) = ab - a c

a = (x^{4}), b = - \frac{ln ( x )}{4 x ^{3}}, c = \frac{1}{12 x ^{3}}

= (x^{4}) (- \frac{ln ( x )}{4 x ^{3}}) - (x^{4}) \frac{1}{12 x ^{3}}

Apply minus-plus rules

+ (- a) = - a

= - \frac{ln ( x )}{4 x ^{3}} (x^{4}) - \frac{1}{12 x ^{3}} (x^{4})

Simplify

- \frac{ln ( x )}{4 x ^{3}} (x^{4}) - \frac{1}{12 x ^{3}} (x^{4}) : - \frac{x ln ( x )}{4} - \frac{x}{12}

= - \frac{x ln ( x )}{4} - \frac{x}{12}

(x - \frac{5}{8}) (x + \frac{5}{8})

(x + 3) (x + 3)^{'}

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

1 - cos (x) + ln (2) (1 + tan (22 x)) + 2 arcsin (3 x)

\frac{e ^{2} π}{π ^{2} + 1} cos (π (t - 2))

What is expand (x^4)(-(ln(x))/(4x^3)-1/(12x^3)) ?
The solution to expand (x^4)(-(ln(x))/(4x^3)-1/(12x^3)) is -(xln(x))/4-x/(12)