What is (\partial)/(\partial x)(3x^2-y^3+x-1) ?

The answer to (\partial)/(\partial x)(3x^2-y^3+x-1) is 6x+1

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(\partial)/(\partial x)(3x^2-y^3+x-1)

\frac{\partial}{\partial x} (3 x^{2} - y^{3} + x - 1)

6 x + 1

Solution steps

\frac{\partial}{\partial x} (3 x^{2} - y^{3} + x - 1)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (3 x^{2}) - \frac{\partial}{\partial x} (y^{3}) + \frac{\partial}{\partial x} (x) - \frac{\partial}{\partial x} (1)

\frac{\partial}{\partial x} (3 x^{2}) = 6 x

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

\frac{\partial}{\partial x} (x) = 1

\frac{\partial}{\partial x} (1) = 0

= 6 x - 0 + 1 - 0

Simplify

= 6 x + 1

\int \frac{( ln ( x ^{8} ) )}{x} d x

\int sin (\frac{1}{3}) x d x

\frac{\partial}{\partial x} (9 y e^{x} - ln (z))

\frac{s + 2}{s ^{2} + 2 s + 5}

(t - 2) e^{- 4 t}

What is (\partial)/(\partial x)(3x^2-y^3+x-1) ?
The answer to (\partial)/(\partial x)(3x^2-y^3+x-1) is 6x+1