What is (\partial)/(\partial y)(xzcos(x+yz)) ?

The answer to (\partial)/(\partial y)(xzcos(x+yz)) is -xz^2sin(x+zy)

Popular >

(\partial)/(\partial y)(xzcos(x+yz))

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

- x z^{2} sin (x + zy)

Solution steps

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

Treat

x, z

as constants

Take the constant out:

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

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

Apply the chain rule:

- sin (x + yz) \frac{\partial}{\partial y} (x + yz)

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

\frac{\partial}{\partial y} (x + yz) = z

= x z (- sin (x + yz) z)

Simplify

x z (- sin (x + yz) z) : - x z^{2} sin (x + zy)

= - x z^{2} sin (x + zy)

\frac{d}{d x} (sin^{2} (ln (x)))

\frac{3 x}{9 x ^{2} + 4 x - 9}

x \to 1 lim (\frac{[ x ]}{∣ x ∣})

x \to π lim (sin (6 x))

What is (\partial)/(\partial y)(xzcos(x+yz)) ?
The answer to (\partial)/(\partial y)(xzcos(x+yz)) is -xz^2sin(x+zy)