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

The answer to (\partial)/(\partial x)(4x^2-x) is 8x-1

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(\partial)/(\partial x)(4x^2-x)

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

8 x - 1

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (4 x^{2}) = 8 x

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

= 8 x - 1

f (x) = x + 2, at x = 3

\int_{0}^{\frac{π}{4}} 7 x sec^{2} (x) d x

\int_{- 1}^{0} x e^{x} d x

\frac{d}{d x} (x)

\int 2 (2 x + 1)^{6} d x

What is (\partial)/(\partial x)(4x^2-x) ?
The answer to (\partial)/(\partial x)(4x^2-x) is 8x-1