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

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

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

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

- sin (y + z)

Solution steps

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

Treat

x, z

as constants

Apply the Sum/Difference Rule:

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

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

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

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

= 0 - sin (y + z)

Simplify

= - sin (y + z)

(x^{2} + 4) y^{^{'}} + 3 x y = x

n = 1 \sum \infty \frac{7}{4 ^{n}}

x \to 0 lim (\frac{e ^{a x} - 1}{x})

x - a x

sin^{4} (a) cos^{9} (a)

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