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

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

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

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

2 x - 3 y

Solution steps

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

Treat

y, z

as constants

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (3 x y) = 3 y

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

\frac{\partial}{\partial x} (z^{3}) = 0

= 2 x - 3 y + 0 + 0

Simplify

= 2 x - 3 y

x \to \infty lim (1 + \frac{7}{x})

\frac{s}{( s - 3 ) ^{3}}

\frac{( x ^{3} )}{3}

f (x) = 9 x^{3} - 7 x

y^{^{''}} = (y^{^{'}})^{2} + 1

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