What is the derivative of-9(3x^4-x^3+2x^5) ?

The derivative of-9(3x^4-x^3+2x^5) is -45(3x^4-x^3+2x)^4(12x^3-3x^2+2)

What is the first derivative of-9(3x^4-x^3+2x^5) ?

The first derivative of-9(3x^4-x^3+2x^5) is -45(3x^4-x^3+2x)^4(12x^3-3x^2+2)

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\frac{d}{d x} (- 9 (3 x^{4} - x^{3} + 2 x)^{5})

- 45 (3 x^{4} - x^{3} + 2 x)^{4} (12 x^{3} - 3 x^{2} + 2)

Solution steps

\frac{d}{d x} (- 9 (3 x^{4} - x^{3} + 2 x)^{5})

Take the constant out:

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

= - 9 \frac{d}{d x} ((3 x^{4} - x^{3} + 2 x)^{5})

Apply the chain rule:

5 (3 x^{4} - x^{3} + 2 x)^{4} \frac{d}{d x} (3 x^{4} - x^{3} + 2 x)

= 5 (3 x^{4} - x^{3} + 2 x)^{4} \frac{d}{d x} (3 x^{4} - x^{3} + 2 x)

\frac{d}{d x} (3 x^{4} - x^{3} + 2 x) = 12 x^{3} - 3 x^{2} + 2

= - 9 \cdot 5 (3 x^{4} - x^{3} + 2 x)^{4} (12 x^{3} - 3 x^{2} + 2)

Simplify

= - 45 (3 x^{4} - x^{3} + 2 x)^{4} (12 x^{3} - 3 x^{2} + 2)

5 x^{2} + 2 x y - 3 y^{2} + 3 = 0, (2, 1)

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

\frac{\partial}{\partial y} (2 x - 3 y + 5)

(x^{2} + 2) (x^{3} + 6)

\int cos (x) e^{- s x} d x

What is the derivative of-9(3x^4-x^3+2x^5) ?
The derivative of-9(3x^4-x^3+2x^5) is -45(3x^4-x^3+2x)^4(12x^3-3x^2+2)
What is the first derivative of-9(3x^4-x^3+2x^5) ?
The first derivative of-9(3x^4-x^3+2x^5) is -45(3x^4-x^3+2x)^4(12x^3-3x^2+2)