What is the derivative of (x^3/3+7x^2+49x) ?

The derivative of (x^3/3+7x^2+49x) is x^2+14x+49

What is the first derivative of (x^3/3+7x^2+49x) ?

The first derivative of (x^3/3+7x^2+49x) is x^2+14x+49

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\frac{d}{d x} (\frac{x ^{3}}{3} + 7 x^{2} + 49 x)

x^{2} + 14 x + 49

Solution steps

\frac{d}{d x} (\frac{x ^{3}}{3} + 7 x^{2} + 49 x)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{x ^{3}}{3}) + \frac{d}{d x} (7 x^{2}) + \frac{d}{d x} (49 x)

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

\frac{d}{d x} (7 x^{2}) = 14 x

\frac{d}{d x} (49 x) = 49

= x^{2} + 14 x + 49

x \to m lim (x - [x])

\int 5 r e^{\frac{r}{2}} d r

f (x) = x^{2} ln (8 - 5 x^{2})

y = x^{3} - 10 x, y = 6 x

\frac{d y}{d x} = x^{6} y^{- 5}

What is the derivative of (x^3/3+7x^2+49x) ?
The derivative of (x^3/3+7x^2+49x) is x^2+14x+49
What is the first derivative of (x^3/3+7x^2+49x) ?
The first derivative of (x^3/3+7x^2+49x) is x^2+14x+49