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

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

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

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

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derivative of

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

- \frac{2}{x}

Solution steps

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

Simplify

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

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

= - \frac{1}{x ^{2}} \cdot 2 x

Simplify

- \frac{1}{x ^{2}} \cdot 2 x : - \frac{2}{x}

= - \frac{2}{x}

\frac{5}{3}

y^{^{'}} = 5 x y

\frac{d}{d x} (sin^{\frac{2}{3}} (sec (tan^{2} (2 x))))

- 3 x^{3} - 2 x^{2} - x + 3

x^{2}, 2 x - 1, 0

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