What is the derivative of f(x)=x^4+4x^3-48x^2+6 ?

The derivative of f(x)=x^4+4x^3-48x^2+6 is 4x^3+12x^2-96x

What is the first derivative of f(x)=x^4+4x^3-48x^2+6 ?

The first derivative of f(x)=x^4+4x^3-48x^2+6 is 4x^3+12x^2-96x

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derivative of

f (x) = x^{4} + 4 x^{3} - 48 x^{2} + 6

4 x^{3} + 12 x^{2} - 96 x

Solution steps

\frac{d}{d x} (x^{4} + 4 x^{3} - 48 x^{2} + 6)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (x^{4}) + \frac{d}{d x} (4 x^{3}) - \frac{d}{d x} (48 x^{2}) + \frac{d}{d x} (6)

\frac{d}{d x} (x^{4}) = 4 x^{3}

\frac{d}{d x} (4 x^{3}) = 12 x^{2}

\frac{d}{d x} (48 x^{2}) = 96 x

\frac{d}{d x} (6) = 0

= 4 x^{3} + 12 x^{2} - 96 x + 0

Simplify

= 4 x^{3} + 12 x^{2} - 96 x

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- 4 x^{2} + 12 x, 2 x^{2} - 5

x \to - 3 lim (10 x - 4)

\frac{d}{d x} (ln (7 - x))

What is the derivative of f(x)=x^4+4x^3-48x^2+6 ?
The derivative of f(x)=x^4+4x^3-48x^2+6 is 4x^3+12x^2-96x
What is the first derivative of f(x)=x^4+4x^3-48x^2+6 ?
The first derivative of f(x)=x^4+4x^3-48x^2+6 is 4x^3+12x^2-96x