What is the derivative of x^3+4x^2 ?

The derivative of x^3+4x^2 is 3x^2+8x

What is the first derivative of x^3+4x^2 ?

The first derivative of x^3+4x^2 is 3x^2+8x

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

3 x^{2} + 8 x

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d}{d x} (4 x^{2}) = 8 x

= 3 x^{2} + 8 x

y^{^{'}} = \frac{x ^{4} y ^{4} + x ^{4}}{y ^{3}}

\frac{d}{d x} (\frac{3}{x + 3})

x \to 0 + lim (x^{16 x})

y^{^{'''}} + 2 x \cdot x = 0

\int \frac{x ^{5} + x ^{4} - 8}{x ^{3} - 4 x} d x

What is the derivative of x^3+4x^2 ?
The derivative of x^3+4x^2 is 3x^2+8x
What is the first derivative of x^3+4x^2 ?
The first derivative of x^3+4x^2 is 3x^2+8x