What is (\partial)/(\partial x)(x^3-7x^2+y^3-5y) ?

The answer to (\partial)/(\partial x)(x^3-7x^2+y^3-5y) is 3x^2-14x

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(\partial)/(\partial x)(x^3-7x^2+y^3-5y)

\frac{\partial}{\partial x} (x^{3} - 7 x^{2} + y^{3} - 5 y)

3 x^{2} - 14 x

Solution steps

\frac{\partial}{\partial x} (x^{3} - 7 x^{2} + y^{3} - 5 y)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (x^{3}) - \frac{\partial}{\partial x} (7 x^{2}) + \frac{\partial}{\partial x} (y^{3}) - \frac{\partial}{\partial x} (5 y)

\frac{\partial}{\partial x} (x^{3}) = 3 x^{2}

\frac{\partial}{\partial x} (7 x^{2}) = 14 x

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

\frac{\partial}{\partial x} (5 y) = 0

= 3 x^{2} - 14 x + 0 - 0

Simplify

= 3 x^{2} - 14 x

5 + 3 x

x \to 0 - lim (\frac{ln ( x )}{x})

\int 7 tan (x) sec^{3} (x) d x

\int_{0}^{6} (6 x^{2} - 1) d x

2 x ln (1 + \frac{x ^{2}}{4})

What is (\partial)/(\partial x)(x^3-7x^2+y^3-5y) ?
The answer to (\partial)/(\partial x)(x^3-7x^2+y^3-5y) is 3x^2-14x