What is the first derivative of 2x^2-3 ?

The first derivative of 2x^2-3 is 4x

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derivative of 2x^2-3

derivative of

2 x^{2} - 3

4 x

Solution steps

\frac{d}{d x} (2 x^{2} - 3)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (2 x^{2}) - \frac{d}{d x} (3)

\frac{d}{d x} (2 x^{2}) = 4 x

\frac{d}{d x} (3) = 0

= 4 x - 0

Simplify

= 4 x

\int \frac{1}{25 x ^{2} - 4} d x

\frac{d}{d x} (\frac{1}{2} x^{3} + \frac{1}{6 x})

\frac{\partial}{\partial x} (3^{4} - 6 5^{3})

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

5 cos (7 x), 5 - 5 cos (7 x), [0, \frac{π}{7}]