What is the derivative of y=ln(1/((x-2)^3)) ?

The derivative of y=ln(1/((x-2)^3)) is -3/(x-2)

What is the first derivative of y=ln(1/((x-2)^3)) ?

The first derivative of y=ln(1/((x-2)^3)) is -3/(x-2)

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derivative of

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

- \frac{3}{x - 2}

Solution steps

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

Simplify

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

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

= - \frac{1}{( x - 2 ) ^{3}} \cdot 3 (x - 2)^{2}

Simplify

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

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

- 5 e^{4 t}

\frac{\partial}{\partial x} (x sin (2 x^{2} y))

\int \frac{1}{t 4 t ^{2} - 1} d t

y^{^{''}} - y = x^{2} e^{x}

\int 2 (x - 1) d x

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