What is the derivative of (x^4/8 ln(x)-(x^4)/(32)) ?

The derivative of (x^4/8 ln(x)-(x^4)/(32)) is (x^3ln(x))/2

What is the first derivative of (x^4/8 ln(x)-(x^4)/(32)) ?

The first derivative of (x^4/8 ln(x)-(x^4)/(32)) is (x^3ln(x))/2

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\frac{d}{d x} (\frac{x ^{4}}{8} ln (x) - \frac{x ^{4}}{32})

\frac{x ^{3} ln ( x )}{2}

Solution steps

\frac{d}{d x} (\frac{x ^{4}}{8} ln (x) - \frac{x ^{4}}{32})

Simplify

\frac{x ^{4}}{8} ln (x) - \frac{x ^{4}}{32} : \frac{x ^{4} ln ( x )}{8} - \frac{x ^{4}}{32}

= \frac{d}{d x} (\frac{x ^{4} ln ( x )}{8} - \frac{x ^{4}}{32})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (\frac{x ^{4} ln ( x )}{8}) - \frac{d}{d x} (\frac{x ^{4}}{32})

\frac{d}{d x} (\frac{x ^{4} ln ( x )}{8}) = \frac{1}{8} (4 x^{3} ln (x) + x^{3})

\frac{d}{d x} (\frac{x ^{4}}{32}) = \frac{x ^{3}}{8}

= \frac{1}{8} (4 x^{3} ln (x) + x^{3}) - \frac{x ^{3}}{8}

Simplify

\frac{1}{8} (4 x^{3} ln (x) + x^{3}) - \frac{x ^{3}}{8} : \frac{x ^{3} ln ( x )}{2}

= \frac{x ^{3} ln ( x )}{2}

y = \frac{8}{x + 1}

x \to 0 lim (x \cdot e^{\frac{1}{x}})

y = (x^{4} - 6 x) e^{x}

\frac{\partial}{\partial x} (\frac{2 x}{x ^{2} + y ^{2} + 16})

\frac{d}{d x} (\frac{cos ( 2 x )}{x})

What is the derivative of (x^4/8 ln(x)-(x^4)/(32)) ?
The derivative of (x^4/8 ln(x)-(x^4)/(32)) is (x^3ln(x))/2
What is the first derivative of (x^4/8 ln(x)-(x^4)/(32)) ?
The first derivative of (x^4/8 ln(x)-(x^4)/(32)) is (x^3ln(x))/2