What is the derivative of 400(15-x^2(3x-2)) ?

The derivative of 400(15-x^2(3x-2)) is 400(-9x^2+4x+45)

What is the first derivative of 400(15-x^2(3x-2)) ?

The first derivative of 400(15-x^2(3x-2)) is 400(-9x^2+4x+45)

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\frac{d}{d x} (400 (15 - x^{2}) (3 x - 2))

400 (- 9 x^{2} + 4 x + 45)

Solution steps

\frac{d}{d x} (400 (15 - x^{2}) (3 x - 2))

Take the constant out:

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

= 400 \frac{d}{d x} ((15 - x^{2}) (3 x - 2))

Apply the Product Rule:

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

f = 15 - x^{2}, g = 3 x - 2

= 400 (\frac{d}{d x} (15 - x^{2}) (3 x - 2) + \frac{d}{d x} (3 x - 2) (15 - x^{2}))

\frac{d}{d x} (15 - x^{2}) = - 2 x

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

= 400 ((- 2 x) (3 x - 2) + 3 (15 - x^{2}))

Simplify

400 ((- 2 x) (3 x - 2) + 3 (15 - x^{2})) : 400 (- 9 x^{2} + 4 x + 45)

= 400 (- 9 x^{2} + 4 x + 45)

What is the derivative of 400(15-x^2(3x-2)) ?
The derivative of 400(15-x^2(3x-2)) is 400(-9x^2+4x+45)
What is the first derivative of 400(15-x^2(3x-2)) ?
The first derivative of 400(15-x^2(3x-2)) is 400(-9x^2+4x+45)