What is the derivative of 4x^4+1/3 x^3+x+1 ?

The derivative of 4x^4+1/3 x^3+x+1 is 16x^3+x^2+1

What is the first derivative of 4x^4+1/3 x^3+x+1 ?

The first derivative of 4x^4+1/3 x^3+x+1 is 16x^3+x^2+1

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\frac{d}{d x} (4 x^{4} + \frac{1}{3} x^{3} + x + 1)

16 x^{3} + x^{2} + 1

Solution steps

\frac{d}{d x} (4 x^{4} + \frac{1}{3} x^{3} + x + 1)

Simplify

4 x^{4} + \frac{1}{3} x^{3} + x + 1 : 4 x^{4} + \frac{x ^{3}}{3} + x + 1

= \frac{d}{d x} (4 x^{4} + \frac{x ^{3}}{3} + x + 1)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (4 x^{4}) + \frac{d}{d x} (\frac{x ^{3}}{3}) + \frac{d x}{d x} + \frac{d}{d x} (1)

\frac{d}{d x} (4 x^{4}) = 16 x^{3}

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

\frac{d x}{d x} = 1

\frac{d}{d x} (1) = 0

= 16 x^{3} + x^{2} + 1 + 0

Simplify

= 16 x^{3} + x^{2} + 1

What is the derivative of 4x^4+1/3 x^3+x+1 ?
The derivative of 4x^4+1/3 x^3+x+1 is 16x^3+x^2+1
What is the first derivative of 4x^4+1/3 x^3+x+1 ?
The first derivative of 4x^4+1/3 x^3+x+1 is 16x^3+x^2+1