What is the first derivative of 4x^2+1 ?

The first derivative of 4x^2+1 is 8x

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derivative of 4x^2+1

derivative of

4 x^{2} + 1

8 x

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

= 8 x + 0

Simplify

= 8 x

4 y^{^{''}} - 4 y^{^{'}} + 10 y = 0

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

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

x \to 6 lim (\frac{x ^{2}}{x - 6})

\frac{d y}{d x} + y^{5} x = - 6 y