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

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

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

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

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derivative of

f (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}}

f (x) = tan (x)

(3, 4)

θ = \frac{5 π}{3}

y = (cos (x))^{x}

(3, - 2)

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