What is (\partial)/(\partial z)(4x-yz+3yz^2+z^2) ?

The answer to (\partial)/(\partial z)(4x-yz+3yz^2+z^2) is 6yz-y+2z

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(\partial)/(\partial z)(4x-yz+3yz^2+z^2)

\frac{\partial}{\partial z} (4 x - yz + 3 y z^{2} + z^{2})

6 yz - y + 2 z

Solution steps

\frac{\partial}{\partial z} (4 x - yz + 3 y z^{2} + z^{2})

Treat

x, y

as constants

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial z} (4 x) - \frac{\partial}{\partial z} (yz) + \frac{\partial}{\partial z} (3 y z^{2}) + \frac{\partial}{\partial z} (z^{2})

\frac{\partial}{\partial z} (4 x) = 0

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

\frac{\partial}{\partial z} (3 y z^{2}) = 6 yz

\frac{\partial}{\partial z} (z^{2}) = 2 z

= 0 - y + 6 yz + 2 z

Simplify

= 6 yz - y + 2 z

f (x) = sin (x), - \frac{2 π}{3}, \frac{π}{3}

\int \frac{6 x}{( x ^{2} + 3 ) ^{2}} d x

3 - x^{2}, 11 - 6 x

\int 8 x cos (x) d x

\int \frac{3 x + 2}{x ^{2} - 11} d x

What is (\partial)/(\partial z)(4x-yz+3yz^2+z^2) ?
The answer to (\partial)/(\partial z)(4x-yz+3yz^2+z^2) is 6yz-y+2z