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

The answer to (\partial)/(\partial y)((cos(y))/(xz^2)) is -(sin(y))/(xz^2)

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

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

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

Solution steps

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

Treat

x, z

as constants

Take the constant out:

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

= \frac{1}{x z ^{2}} \frac{\partial}{\partial y} (cos (y))

Apply the common derivative:

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

= \frac{1}{x z ^{2}} (- sin (y))

Simplify

\frac{1}{x z ^{2}} (- sin (y)) : - \frac{sin ( y )}{x z ^{2}}

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

y^{^{'''}} = 6

\int sin (3 π t) d t

\int \frac{1}{5 - 2 x} d x

6 x - 4 x^{3} + 9

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

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