What is (\partial)/(\partial y)(2x-3yz) ?

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

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(\partial)/(\partial y)(2x-3yz)

\frac{\partial}{\partial y} (2 x - 3 yz)

- 3 z

Solution steps

\frac{\partial}{\partial y} (2 x - 3 yz)

Treat

x, z

as constants

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (2 x) - \frac{\partial}{\partial y} (3 yz)

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

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

= 0 - 3 z

Simplify

= - 3 z

\frac{d}{d r} (\frac{200 r ( r + 2 )}{( 2 + 3 r ) ^{2}})

y = 9 - 10 x, at x = - 1

\frac{d}{d x} (2 x y^{3})

\frac{\partial}{\partial y} (e^{s i n (x + y)})

f (x) = 1 + 4 csc (x), at x = - \frac{π}{3}

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