What is the derivative of 100+0.4x^2 ?

The derivative of 100+0.4x^2 is 0.8x

What is the first derivative of 100+0.4x^2 ?

The first derivative of 100+0.4x^2 is 0.8x

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derivative of

100 + 0.4 x^{2}

0.8 x

Solution steps

\frac{d}{d x} (100 + 0.4 x^{2})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (100) + \frac{d}{d x} (0.4 x^{2})

\frac{d}{d x} (100) = 0

\frac{d}{d x} (0.4 x^{2}) = 0.8 x

= 0 + 0.8 x

Simplify

= 0.8 x

d er i v a t i v e f (x) = 8 + x tan (x)

t an g e n t f (x) = 3 e^{x} cos (x), (0, 3)

\int 2 e^{- 2 t} + 5 e^{- 8 t} d t

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

x \to 1 - lim (4 x^{3} - e^{x - 1})

What is the derivative of 100+0.4x^2 ?
The derivative of 100+0.4x^2 is 0.8x
What is the first derivative of 100+0.4x^2 ?
The first derivative of 100+0.4x^2 is 0.8x