What is the derivative of 64x-0.02x^2 ?

The derivative of 64x-0.02x^2 is 64-0.04x

What is the first derivative of 64x-0.02x^2 ?

The first derivative of 64x-0.02x^2 is 64-0.04x

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derivative of

64 x - 0.02 x^{2}

64 - 0.04 x

Solution steps

\frac{d}{d x} (64 x - 0.02 x^{2})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (64 x) - \frac{d}{d x} (0.02 x^{2})

\frac{d}{d x} (64 x) = 64

\frac{d}{d x} (0.02 x^{2}) = 0.04 x

= 64 - 0.04 x

f (x) = 3 x^{2} + x, at x = 1

sin^{2} (t)

\frac{d}{d x} (\frac{x}{2 x + 3})

\int_{1}^{1 - 2 y} \frac{y}{x} d x

\frac{d}{d x} (9 x^{2} + y^{2})

What is the derivative of 64x-0.02x^2 ?
The derivative of 64x-0.02x^2 is 64-0.04x
What is the first derivative of 64x-0.02x^2 ?
The first derivative of 64x-0.02x^2 is 64-0.04x