What is (\partial)/(\partial y)(e^x+cos(y)) ?

The answer to (\partial)/(\partial y)(e^x+cos(y)) is -sin(y)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial y)(e^x+cos(y))

\frac{\partial}{\partial y} (e^{x} + cos (y))

- sin (y)

Solution steps

\frac{\partial}{\partial y} (e^{x} + cos (y))

Treat

x

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial y} (e^{x}) + \frac{\partial}{\partial y} (cos (y))

\frac{\partial}{\partial y} (e^{x}) = 0

\frac{\partial}{\partial y} (cos (y)) = - sin (y)

= 0 - sin (y)

Simplify

= - sin (y)

ln (8)

\int \frac{1}{x ln ( x ) - x} d x

\frac{\partial}{\partial x} (x e^{y} - e^{z})

\frac{d y}{d x}, y e^{x} = y

\frac{d}{d x} ((3 x + 1) \frac{( x ^{2} + 4 )}{( 6 x - 5 )})

What is (\partial)/(\partial y)(e^x+cos(y)) ?
The answer to (\partial)/(\partial y)(e^x+cos(y)) is -sin(y)