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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{- 3}, g = ln (x + x^{- 1})

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

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

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

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

Simplify

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

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

x \to - 9 lim (\frac{x ^{2} - 7}{9 - x})

\frac{\partial}{\partial x} (4 x^{2} y^{3} + 7 x y^{2} - 7 x^{2})

y = 4 π^{2}

\int (x^{2} + 3 x)^{12} (2 x + 3) d x

250 y^{^{''}} + 8 y^{^{'}} + 2 y = 0

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