What is (\partial)/(\partial x)(x^8+y^9+z^3-9xyz) ?

The answer to (\partial)/(\partial x)(x^8+y^9+z^3-9xyz) is 8x^7-9yz

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(\partial)/(\partial x)(x^8+y^9+z^3-9xyz)

\frac{\partial}{\partial x} (x^{8} + y^{9} + z^{3} - 9 x yz)

8 x^{7} - 9 yz

Solution steps

\frac{\partial}{\partial x} (x^{8} + y^{9} + z^{3} - 9 x yz)

Treat

y, z

as constants

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (x^{8}) + \frac{\partial}{\partial x} (y^{9}) + \frac{\partial}{\partial x} (z^{3}) - \frac{\partial}{\partial x} (9 x yz)

\frac{\partial}{\partial x} (x^{8}) = 8 x^{7}

\frac{\partial}{\partial x} (y^{9}) = 0

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

\frac{\partial}{\partial x} (9 x yz) = 9 yz

= 8 x^{7} + 0 + 0 - 9 yz

Simplify

= 8 x^{7} - 9 yz

x^{3} y^{^{'''}} + x y^{^{'}} - y = x^{2}

\int 20 x^{3} (5 x^{4} + 6)^{3} d x

x \to 4 - lim (\frac{x}{x - 2})

\frac{1}{x}, 1, 33

\int \frac{42 x ^{2}}{x ^{4} - 29 x ^{2} + 100} d x

What is (\partial)/(\partial x)(x^8+y^9+z^3-9xyz) ?
The answer to (\partial)/(\partial x)(x^8+y^9+z^3-9xyz) is 8x^7-9yz