What is the derivative of sin(-2ln(1-2x)) ?

The derivative of sin(-2ln(1-2x)) is (4cos(2ln(1-2x)))/(1-2x)

What is the first derivative of sin(-2ln(1-2x)) ?

The first derivative of sin(-2ln(1-2x)) is (4cos(2ln(1-2x)))/(1-2x)

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

\frac{4 cos ( 2 ln ( 1 - 2 x ) )}{1 - 2 x}

Solution steps

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

Simplify

sin (- 2 ln (1 - 2 x)) : - sin (2 ln (1 - 2 x))

= \frac{d}{d x} (- sin (2 ln (1 - 2 x)))

Take the constant out:

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

= - \frac{d}{d x} (sin (2 ln (1 - 2 x)))

Apply the chain rule:

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

= cos (2 ln (1 - 2 x)) \frac{d}{d x} (2 ln (1 - 2 x))

\frac{d}{d x} (2 ln (1 - 2 x)) = - \frac{4}{1 - 2 x}

= - cos (2 ln (1 - 2 x)) (- \frac{4}{1 - 2 x})

Simplify

- cos (2 ln (1 - 2 x)) (- \frac{4}{1 - 2 x}) : \frac{4 cos ( 2 ln ( 1 - 2 x ) )}{1 - 2 x}

= \frac{4 cos ( 2 ln ( 1 - 2 x ) )}{1 - 2 x}

What is the derivative of sin(-2ln(1-2x)) ?
The derivative of sin(-2ln(1-2x)) is (4cos(2ln(1-2x)))/(1-2x)
What is the first derivative of sin(-2ln(1-2x)) ?
The first derivative of sin(-2ln(1-2x)) is (4cos(2ln(1-2x)))/(1-2x)