What is the derivative of 16x^3-96x^2+178x-95 ?

The derivative of 16x^3-96x^2+178x-95 is 48x^2-192x+178

What is the first derivative of 16x^3-96x^2+178x-95 ?

The first derivative of 16x^3-96x^2+178x-95 is 48x^2-192x+178

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\frac{d}{d x} (16 x^{3} - 96 x^{2} + 178 x - 95)

48 x^{2} - 192 x + 178

Solution steps

\frac{d}{d x} (16 x^{3} - 96 x^{2} + 178 x - 95)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (16 x^{3}) - \frac{d}{d x} (96 x^{2}) + \frac{d}{d x} (178 x) - \frac{d}{d x} (95)

\frac{d}{d x} (16 x^{3}) = 48 x^{2}

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

\frac{d}{d x} (178 x) = 178

\frac{d}{d x} (95) = 0

= 48 x^{2} - 192 x + 178 - 0

Simplify

= 48 x^{2} - 192 x + 178

What is the derivative of 16x^3-96x^2+178x-95 ?
The derivative of 16x^3-96x^2+178x-95 is 48x^2-192x+178
What is the first derivative of 16x^3-96x^2+178x-95 ?
The first derivative of 16x^3-96x^2+178x-95 is 48x^2-192x+178