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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

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

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

\frac{d}{d x} (4 x^{2} + x) = 8 x + 1

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

Simplify

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

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

y^{^{'}} = - 3 y

\frac{\partial}{\partial x} (e^{6 x + 4 y})

\int_{0}^{1} \frac{4}{4 - x ^{2}} d x

y = \frac{2 x + 5}{5 x ^{2}}

\int \frac{48 x ^{2}}{( x - 9 ) ( x + 3 ) ^{2}} d x

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