What is the derivative of f(v)=(v^{-3}+6v^{-2})^3 ?

The derivative of f(v)=(v^{-3}+6v^{-2})^3 is 3(1/(v^3)+6/(v^2))^2(-3/(v^4)-(12)/(v^3))

What is the first derivative of f(v)=(v^{-3}+6v^{-2})^3 ?

The first derivative of f(v)=(v^{-3}+6v^{-2})^3 is 3(1/(v^3)+6/(v^2))^2(-3/(v^4)-(12)/(v^3))

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derivative of

f (v) = (v^{- 3} + 6 v^{- 2})^{3}

3 (\frac{1}{v ^{3}} + \frac{6}{v ^{2}})^{2} (- \frac{3}{v ^{4}} - \frac{12}{v ^{3}})

Solution steps

\frac{d}{d v} ((v^{- 3} + 6 v^{- 2})^{3})

Apply the chain rule:

3 (v^{- 3} + 6 v^{- 2})^{2} \frac{d}{d v} (v^{- 3} + 6 v^{- 2})

= 3 (v^{- 3} + 6 v^{- 2})^{2} \frac{d}{d v} (v^{- 3} + 6 v^{- 2})

\frac{d}{d v} (v^{- 3} + 6 v^{- 2}) = - \frac{3}{v ^{4}} - \frac{12}{v ^{3}}

= 3 (v^{- 3} + 6 v^{- 2})^{2} (- \frac{3}{v ^{4}} - \frac{12}{v ^{3}})

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

= 3 (\frac{1}{v ^{3}} + \frac{6}{v ^{2}})^{2} (- \frac{3}{v ^{4}} - \frac{12}{v ^{3}})

f (x) = \frac{3}{x + 1}

\frac{\partial}{\partial z} (x y cos (x + yz))

\frac{d y}{d x} = 20 - 5 y

\int \frac{2 - 8 x}{1 - 4 x ^{2}} d x

x \to 1 lim (3 x^{3} + 2 x^{2} + 4)

What is the derivative of f(v)=(v^{-3}+6v^{-2})^3 ?
The derivative of f(v)=(v^{-3}+6v^{-2})^3 is 3(1/(v^3)+6/(v^2))^2(-3/(v^4)-(12)/(v^3))
What is the first derivative of f(v)=(v^{-3}+6v^{-2})^3 ?
The first derivative of f(v)=(v^{-3}+6v^{-2})^3 is 3(1/(v^3)+6/(v^2))^2(-3/(v^4)-(12)/(v^3))