What is the derivative of f(x)=(x+9)(x^2-9) ?

The derivative of f(x)=(x+9)(x^2-9) is 3x^2+18x-9

What is the first derivative of f(x)=(x+9)(x^2-9) ?

The first derivative of f(x)=(x+9)(x^2-9) is 3x^2+18x-9

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derivative of

f (x) = (x + 9) (x^{2} - 9)

3 x^{2} + 18 x - 9

Solution steps

\frac{d}{d x} ((x + 9) (x^{2} - 9))

Apply the Product Rule:

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

f = x + 9, g = x^{2} - 9

= \frac{d}{d x} (x + 9) (x^{2} - 9) + \frac{d}{d x} (x^{2} - 9) (x + 9)

\frac{d}{d x} (x + 9) = 1

\frac{d}{d x} (x^{2} - 9) = 2 x

= 1 \cdot (x^{2} - 9) + 2 x (x + 9)

Simplify

1 \cdot (x^{2} - 9) + 2 x (x + 9) : 3 x^{2} + 18 x - 9

= 3 x^{2} + 18 x - 9

\frac{6}{s ^{4} ( s - 2 ) ^{2}}

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

\frac{d}{d x} (2 sec^{2} (5 π x - 1))

\frac{d}{d x} (\frac{5}{x} + \frac{x}{5})

\frac{d y}{d x} = 16 x + y - 16

What is the derivative of f(x)=(x+9)(x^2-9) ?
The derivative of f(x)=(x+9)(x^2-9) is 3x^2+18x-9
What is the first derivative of f(x)=(x+9)(x^2-9) ?
The first derivative of f(x)=(x+9)(x^2-9) is 3x^2+18x-9