What is the derivative of y=96x^2+72x+12 ?

The derivative of y=96x^2+72x+12 is 192x+72

What is the first derivative of y=96x^2+72x+12 ?

The first derivative of y=96x^2+72x+12 is 192x+72

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derivative of

y = 96 x^{2} + 72 x + 12

192 x + 72

Solution steps

\frac{d}{d x} (96 x^{2} + 72 x + 12)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (96 x^{2}) + \frac{d}{d x} (72 x) + \frac{d}{d x} (12)

\frac{d}{d x} (96 x^{2}) = 192 x

\frac{d}{d x} (72 x) = 72

\frac{d}{d x} (12) = 0

= 192 x + 72 + 0

Simplify

= 192 x + 72

What is the derivative of y=96x^2+72x+12 ?
The derivative of y=96x^2+72x+12 is 192x+72
What is the first derivative of y=96x^2+72x+12 ?
The first derivative of y=96x^2+72x+12 is 192x+72