What is the derivative of f(x)=9x^5(x^3-2x) ?

The derivative of f(x)=9x^5(x^3-2x) is 9(8x^7-12x^5)

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

The first derivative of f(x)=9x^5(x^3-2x) is 9(8x^7-12x^5)

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derivative of

f (x) = 9 x^{5} (x^{3} - 2 x)

9 (8 x^{7} - 12 x^{5})

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

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

f = x^{5}, g = x^{3} - 2 x

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

\frac{d}{d x} (x^{5}) = 5 x^{4}

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

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

Expand

5 x^{4} (x^{3} - 2 x) + (3 x^{2} - 2) x^{5} : 8 x^{7} - 12 x^{5}

= 9 (8 x^{7} - 12 x^{5})

y^{^{''}} - 2 y^{^{'}} - 63 y = 0

n = 1 \sum \infty \frac{1}{n ^{- 1}}

\frac{d ^{2} y}{d ^{2} x} + \frac{4 d y}{d x} + 3 y = 0

x \to 0 lim (\frac{x}{x ^{2} - 4})

t^{3} \frac{d y}{d t} + 3 t^{2} y = 2 cos (t)

What is the derivative of f(x)=9x^5(x^3-2x) ?
The derivative of f(x)=9x^5(x^3-2x) is 9(8x^7-12x^5)
What is the first derivative of f(x)=9x^5(x^3-2x) ?
The first derivative of f(x)=9x^5(x^3-2x) is 9(8x^7-12x^5)