What is the derivative of 1/3 x^3+8x^2+63x+7 ?

The derivative of 1/3 x^3+8x^2+63x+7 is x^2+16x+63

What is the first derivative of 1/3 x^3+8x^2+63x+7 ?

The first derivative of 1/3 x^3+8x^2+63x+7 is x^2+16x+63

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derivative of

\frac{1}{3} x^{3} + 8 x^{2} + 63 x + 7

x^{2} + 16 x + 63

Solution steps

\frac{d}{d x} (\frac{1}{3} x^{3} + 8 x^{2} + 63 x + 7)

Simplify

\frac{1}{3} x^{3} + 8 x^{2} + 63 x + 7 : \frac{x ^{3}}{3} + 8 x^{2} + 63 x + 7

= \frac{d}{d x} (\frac{x ^{3}}{3} + 8 x^{2} + 63 x + 7)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{x ^{3}}{3}) + \frac{d}{d x} (8 x^{2}) + \frac{d}{d x} (63 x) + \frac{d}{d x} (7)

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

\frac{d}{d x} (8 x^{2}) = 16 x

\frac{d}{d x} (63 x) = 63

\frac{d}{d x} (7) = 0

= x^{2} + 16 x + 63 + 0

Simplify

= x^{2} + 16 x + 63

\int_{- 4 π}^{4 π} sin^{2} (x) + 1 d x

x \to 2 - lim (4)

y^{^{''}} x + (\frac{2}{x} - \frac{2 x}{1 + x ^{2}}) \cdot y^{^{'}} = 0

\frac{d}{d x} (2 cos (x) - 4 sin (x))

\int 2 arcsin (2 x) d x

What is the derivative of 1/3 x^3+8x^2+63x+7 ?
The derivative of 1/3 x^3+8x^2+63x+7 is x^2+16x+63
What is the first derivative of 1/3 x^3+8x^2+63x+7 ?
The first derivative of 1/3 x^3+8x^2+63x+7 is x^2+16x+63