What is the derivative of sin(sqrt(ln(1-3x))) ?

The derivative of sin(sqrt(ln(1-3x))) is -(3cos(sqrt(ln(1-3x))))/(2sqrt(ln(1-3x))(1-3x))

What is the first derivative of sin(sqrt(ln(1-3x))) ?

The first derivative of sin(sqrt(ln(1-3x))) is -(3cos(sqrt(ln(1-3x))))/(2sqrt(ln(1-3x))(1-3x))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Solution steps

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

Apply the chain rule:

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

= cos (ln (1 - 3 x)) \frac{d}{d x} (ln (1 - 3 x))

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

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

Simplify

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

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

\frac{d}{d t} (\frac{d x}{d t}) + x = sin (2 t)

\int_{0}^{\infty} 8 sin^{2} (a) d a

y^{^{''}} - 10 y^{^{'}} + 25 y = e^{5 x}

\int_{0}^{2} 144 x + 289 d x

16 4 4 x^{4} + 4

What is the derivative of sin(sqrt(ln(1-3x))) ?
The derivative of sin(sqrt(ln(1-3x))) is -(3cos(sqrt(ln(1-3x))))/(2sqrt(ln(1-3x))(1-3x))
What is the first derivative of sin(sqrt(ln(1-3x))) ?
The first derivative of sin(sqrt(ln(1-3x))) is -(3cos(sqrt(ln(1-3x))))/(2sqrt(ln(1-3x))(1-3x))