What is (\partial)/(\partial x)(x^4+3y^2-x) ?

The answer to (\partial)/(\partial x)(x^4+3y^2-x) is 4x^3-1

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(\partial)/(\partial x)(x^4+3y^2-x)

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

4 x^{3} - 1

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (x^{4}) = 4 x^{3}

\frac{\partial}{\partial x} (3 y^{2}) = 0

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

= 4 x^{3} + 0 - 1

Simplify

= 4 x^{3} - 1

\frac{x ^{2} - 4}{x - 2}

\frac{d}{d x} (x + y + 2)

\int \frac{3 x ^{2} + 2 x + 4}{x ^{3} + 4 x} d x

\frac{d r}{d θ} + r tan (θ) = 3 tan (θ)

e^{5} + ln (7)

What is (\partial)/(\partial x)(x^4+3y^2-x) ?
The answer to (\partial)/(\partial x)(x^4+3y^2-x) is 4x^3-1