What is (\partial)/(\partial x)(acos(kx)) ?

The answer to (\partial)/(\partial x)(acos(kx)) is -aksin(kx)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial x)(acos(kx))

\frac{\partial}{\partial x} (a cos (k x))

- ak sin (k x)

Solution steps

\frac{\partial}{\partial x} (a cos (k x))

Take the constant out:

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

= a \frac{\partial}{\partial x} (cos (k x))

Apply the chain rule:

- sin (k x) \frac{\partial}{\partial x} (k x)

= - sin (k x) \frac{\partial}{\partial x} (k x)

\frac{\partial}{\partial x} (k x) = k

= a (- sin (k x) k)

Simplify

= - ak sin (k x)

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

ln (x e^{- 3 x})

f (x) = \frac{9}{x}

\int e^{- x^{2} \frac{1}{2}} d x

\frac{d}{d x} (\frac{2 x + 6 x ^{2}}{( 1 + 6 x ) ^{2}})

What is (\partial)/(\partial x)(acos(kx)) ?
The answer to (\partial)/(\partial x)(acos(kx)) is -aksin(kx)