What is the derivative of f(x)=3x^2+32x+64 ?

The derivative of f(x)=3x^2+32x+64 is 6x+32

What is the first derivative of f(x)=3x^2+32x+64 ?

The first derivative of f(x)=3x^2+32x+64 is 6x+32

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derivative of

f (x) = 3 x^{2} + 32 x + 64

6 x + 32

Solution steps

\frac{d}{d x} (3 x^{2} + 32 x + 64)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (3 x^{2}) + \frac{d}{d x} (32 x) + \frac{d}{d x} (64)

\frac{d}{d x} (3 x^{2}) = 6 x

\frac{d}{d x} (32 x) = 32

\frac{d}{d x} (64) = 0

= 6 x + 32 + 0

Simplify

= 6 x + 32

3 x^{3}, [0, 6]

\int x (x^{2} - 3 x - 5) d x

f (x) = 2^{x}

\frac{\partial}{\partial s} (rs e^{t})

\int_{3}^{8} x^{3} - π x^{2} d x

What is the derivative of f(x)=3x^2+32x+64 ?
The derivative of f(x)=3x^2+32x+64 is 6x+32
What is the first derivative of f(x)=3x^2+32x+64 ?
The first derivative of f(x)=3x^2+32x+64 is 6x+32