What is the derivative of 1/8 x^4-(32)/(x^3) ?

The derivative of 1/8 x^4-(32)/(x^3) is (x^3)/2+(96)/(x^4)

What is the first derivative of 1/8 x^4-(32)/(x^3) ?

The first derivative of 1/8 x^4-(32)/(x^3) is (x^3)/2+(96)/(x^4)

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derivative of

\frac{1}{8} x^{4} - \frac{32}{x ^{3}}

\frac{x ^{3}}{2} + \frac{96}{x ^{4}}

Solution steps

\frac{d}{d x} (\frac{1}{8} x^{4} - \frac{32}{x ^{3}})

Simplify

\frac{1}{8} x^{4} - \frac{32}{x ^{3}} : \frac{x ^{4}}{8} - \frac{32}{x ^{3}}

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (\frac{x ^{4}}{8}) = \frac{x ^{3}}{2}

\frac{d}{d x} (\frac{32}{x ^{3}}) = - \frac{96}{x ^{4}}

= \frac{x ^{3}}{2} - (- \frac{96}{x ^{4}})

Simplify

= \frac{x ^{3}}{2} + \frac{96}{x ^{4}}

\int_{- 4 π}^{4 π} (sin^{2} (x) + 1) d x

x^{3} - 11 x^{2} + 28 x, - x^{3} + 11 x^{2} - 28 x

x \to \infty lim (6 x - 1)

y^{^{'}} = e^{x^{2}}

\frac{d}{d x} (e^{l n (l n (2 x))})

What is the derivative of 1/8 x^4-(32)/(x^3) ?
The derivative of 1/8 x^4-(32)/(x^3) is (x^3)/2+(96)/(x^4)
What is the first derivative of 1/8 x^4-(32)/(x^3) ?
The first derivative of 1/8 x^4-(32)/(x^3) is (x^3)/2+(96)/(x^4)