What is the derivative of 5x^2(x+47) ?

The derivative of 5x^2(x+47) is 5(3x^2+94x)

What is the first derivative of 5x^2(x+47) ?

The first derivative of 5x^2(x+47) is 5(3x^2+94x)

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\frac{d}{d x} (5 x^{2} (x + 47))

5 (3 x^{2} + 94 x)

Solution steps

\frac{d}{d x} (5 x^{2} (x + 47))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 5 \frac{d}{d x} (x^{2} (x + 47))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{2}, g = x + 47

= 5 (\frac{d}{d x} (x^{2}) (x + 47) + \frac{d}{d x} (x + 47) x^{2})

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

\frac{d}{d x} (x + 47) = 1

= 5 (2 x (x + 47) + 1 \cdot x^{2})

Simplify

5 (2 x (x + 47) + 1 \cdot x^{2}) : 5 (3 x^{2} + 94 x)

= 5 (3 x^{2} + 94 x)

x \to 0 lim (\frac{x}{x ^{2}})

n = 0 \sum \infty 8 (\frac{1}{5})^{n}

f (x) = - sin (- 8 x - 3) + 5

\int_{0}^{1} 6 x 5^{x} d x

x \to a lim (1)

What is the derivative of 5x^2(x+47) ?
The derivative of 5x^2(x+47) is 5(3x^2+94x)
What is the first derivative of 5x^2(x+47) ?
The first derivative of 5x^2(x+47) is 5(3x^2+94x)