What is the derivative of ln(sin(x+y)) ?

The derivative of ln(sin(x+y)) is (cos(x+y)(1+\frac{dy)/(dx))}{sin(x+y)}

What is the first derivative of ln(sin(x+y)) ?

The first derivative of ln(sin(x+y)) is (cos(x+y)(1+\frac{dy)/(dx))}{sin(x+y)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d x} (ln (sin (x + y)))

\frac{cos ( x + y ) ( 1 + \frac{d y}{d x} )}{sin ( x + y )}

Solution steps

\frac{d}{d x} (ln (sin (x + y)))

Apply the chain rule:

\frac{1}{sin ( x + y )} \frac{d}{d x} (sin (x + y))

= \frac{1}{sin ( x + y )} \frac{d}{d x} (sin (x + y))

\frac{d}{d x} (sin (x + y)) = cos (x + y) (1 + \frac{d y}{d x})

= \frac{1}{sin ( x + y )} cos (x + y) (1 + \frac{d y}{d x})

Simplify

\frac{1}{sin ( x + y )} cos (x + y) (1 + \frac{d y}{d x}) : \frac{cos ( x + y ) ( 1 + \frac{d y}{d x} )}{sin ( x + y )}

= \frac{cos ( x + y ) ( 1 + \frac{d y}{d x} )}{sin ( x + y )}

f (x) = 3 x + 6

\frac{d}{d A} (\frac{1 + cot ( A )}{csc ( A )})

f (x) = x^{4} - 3 x^{2} + 5, at x = - 1

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

\frac{d}{d t} (t - sin (t))

What is the derivative of ln(sin(x+y)) ?
The derivative of ln(sin(x+y)) is (cos(x+y)(1+\frac{dy)/(dx))}{sin(x+y)}
What is the first derivative of ln(sin(x+y)) ?
The first derivative of ln(sin(x+y)) is (cos(x+y)(1+\frac{dy)/(dx))}{sin(x+y)}