What is the derivative of 1/2 (3x^2+x^{-3}) ?

The derivative of 1/2 (3x^2+x^{-3}) is -(3(6x+1))/(2(3x^2+x)^4)

What is the first derivative of 1/2 (3x^2+x^{-3}) ?

The first derivative of 1/2 (3x^2+x^{-3}) is -(3(6x+1))/(2(3x^2+x)^4)

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

- \frac{3 ( 6 x + 1 )}{2 ( 3 x ^{2} + x ) ^{4}}

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the chain rule:

- \frac{3}{( 3 x ^{2} + x ) ^{4}} \frac{d}{d x} (3 x^{2} + x)

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

\frac{d}{d x} (3 x^{2} + x) = 6 x + 1

= \frac{1}{2} (- \frac{3}{( 3 x ^{2} + x ) ^{4}} (6 x + 1))

Simplify

\frac{1}{2} (- \frac{3}{( 3 x ^{2} + x ) ^{4}} (6 x + 1)) : - \frac{3 ( 6 x + 1 )}{2 ( 3 x ^{2} + x ) ^{4}}

= - \frac{3 ( 6 x + 1 )}{2 ( 3 x ^{2} + x ) ^{4}}

f (x) = x^{4} sin (2 x)

\frac{d}{d x} ((x + 2) (\frac{300}{x} + 4))

y^{^{'}} - 2 y = 4 - t

x \to 0 + lim (\frac{19}{x ^{2}})

(5 x + 4)^{x}

What is the derivative of 1/2 (3x^2+x^{-3}) ?
The derivative of 1/2 (3x^2+x^{-3}) is -(3(6x+1))/(2(3x^2+x)^4)
What is the first derivative of 1/2 (3x^2+x^{-3}) ?
The first derivative of 1/2 (3x^2+x^{-3}) is -(3(6x+1))/(2(3x^2+x)^4)