What is the derivative of 1/(g^2(x)) ?

The derivative of 1/(g^2(x)) is -2/(g^3x)

What is the first derivative of 1/(g^2(x)) ?

The first derivative of 1/(g^2(x)) is -2/(g^3x)

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derivative of

\frac{1}{g ^{2} ( x )}

- \frac{2}{g ^{3} x}

Solution steps

\frac{d}{d g} (\frac{1}{g ^{2} x})

Treat

x

as a constant

Take the constant out:

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

= \frac{1}{x} \frac{d}{d g} (\frac{1}{g ^{2}})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= \frac{1}{x} (- 2 g^{- 2 - 1})

Simplify

\frac{1}{x} (- 2 g^{- 2 - 1}) : - \frac{2}{g ^{3} x}

= - \frac{2}{g ^{3} x}

10 x (x^{2} + 5)^{4}

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\int (4 x (3 x - 2)) d x

\frac{d}{d x} (2 a arccos (2 x))

sin^{4} (6 x^{2})

What is the derivative of 1/(g^2(x)) ?
The derivative of 1/(g^2(x)) is -2/(g^3x)
What is the first derivative of 1/(g^2(x)) ?
The first derivative of 1/(g^2(x)) is -2/(g^3x)