What is (\partial)/(\partial y)(cos(xy)+xz^2) ?

The answer to (\partial)/(\partial y)(cos(xy)+xz^2) is -xsin(xy)

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(\partial)/(\partial y)(cos(xy)+xz^2)

\frac{\partial}{\partial y} (cos (x y) + x z^{2})

- x sin (x y)

Solution steps

\frac{\partial}{\partial y} (cos (x y) + x z^{2})

Treat

x, z

as constants

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (cos (x y)) + \frac{\partial}{\partial y} (x z^{2})

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

\frac{\partial}{\partial y} (x z^{2}) = 0

= - sin (x y) x + 0

Simplify

= - x sin (x y)

in v erse l a pl a ce \frac{s ^{2}}{( s ^{2} + 4 s + 4 ) ^{2}}

f (x) = \frac{4}{x ^{3}}

d er i v a t i v e \frac{x}{x ^{2} - x + 1}

t a y l or \frac{1}{x - 1}, 25

d er i v a t i v e f (x) = x - 8 3 x

What is (\partial)/(\partial y)(cos(xy)+xz^2) ?
The answer to (\partial)/(\partial y)(cos(xy)+xz^2) is -xsin(xy)