What is the derivative of 1/16 x^4+1/(2x^2) ?

The derivative of 1/16 x^4+1/(2x^2) is (x^3)/4-1/(x^3)

What is the first derivative of 1/16 x^4+1/(2x^2) ?

The first derivative of 1/16 x^4+1/(2x^2) is (x^3)/4-1/(x^3)

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\frac{d}{d x} (\frac{1}{16} x^{4} + \frac{1}{2 x ^{2}})

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

Solution steps

\frac{d}{d x} (\frac{1}{16} x^{4} + \frac{1}{2 x ^{2}})

Simplify

\frac{1}{16} x^{4} + \frac{1}{2 x ^{2}} : \frac{x ^{4}}{16} + \frac{1}{2 x ^{2}}

= \frac{d}{d x} (\frac{x ^{4}}{16} + \frac{1}{2 x ^{2}})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{x ^{4}}{16}) + \frac{d}{d x} (\frac{1}{2 x ^{2}})

\frac{d}{d x} (\frac{x ^{4}}{16}) = \frac{x ^{3}}{4}

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

= \frac{x ^{3}}{4} - \frac{1}{x ^{3}}

x \to 3 lim (2 x + 3)

\frac{d}{d x} (r (x) \cdot cos (x))

n = 0 \sum \infty (- \frac{1}{2})^{n - 1}

\frac{d}{d x} (cos (\frac{1}{x ^{2}}))

f (t) = \frac{t}{2}

What is the derivative of 1/16 x^4+1/(2x^2) ?
The derivative of 1/16 x^4+1/(2x^2) is (x^3)/4-1/(x^3)
What is the first derivative of 1/16 x^4+1/(2x^2) ?
The first derivative of 1/16 x^4+1/(2x^2) is (x^3)/4-1/(x^3)