What is the derivative of f(x)=-1/6 (x^{-6}-x^{12}) ?

The derivative of f(x)=-1/6 (x^{-6}-x^{12}) is (2x^{18}+1)/(x^7)

What is the first derivative of f(x)=-1/6 (x^{-6}-x^{12}) ?

The first derivative of f(x)=-1/6 (x^{-6}-x^{12}) is (2x^{18}+1)/(x^7)

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derivative of

f (x) = - \frac{1}{6} (x^{- 6} - x^{12})

\frac{2 x ^{18} + 1}{x ^{7}}

Solution steps

\frac{d}{d x} (- \frac{1}{6} (x^{- 6} - x^{12}))

Take the constant out:

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

= - \frac{1}{6} \frac{d}{d x} (x^{- 6} - x^{12})

Apply the Sum/Difference Rule:

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

= - \frac{1}{6} (\frac{d}{d x} (x^{- 6}) - \frac{d}{d x} (x^{12}))

\frac{d}{d x} (x^{- 6}) = - \frac{6}{x ^{7}}

\frac{d}{d x} (x^{12}) = 12 x^{11}

= - \frac{1}{6} (- \frac{6}{x ^{7}} - 12 x^{11})

Simplify

- \frac{1}{6} (- \frac{6}{x ^{7}} - 12 x^{11}) : \frac{2 x ^{18} + 1}{x ^{7}}

= \frac{2 x ^{18} + 1}{x ^{7}}

x y^{^{'}} + y = x^{\frac{3}{2}} y^{\frac{5}{2}} ln (x)

e^{6 x}

\frac{x + 9}{x}

\frac{d}{d t} (e^{- t} sin (2 t))

(- 4, 0), (1, - 5)

What is the derivative of f(x)=-1/6 (x^{-6}-x^{12}) ?
The derivative of f(x)=-1/6 (x^{-6}-x^{12}) is (2x^{18}+1)/(x^7)
What is the first derivative of f(x)=-1/6 (x^{-6}-x^{12}) ?
The first derivative of f(x)=-1/6 (x^{-6}-x^{12}) is (2x^{18}+1)/(x^7)