What is the derivative of 1/3 x^6(x^2-3) ?

The derivative of 1/3 x^6(x^2-3) is (8x^7-18x^5)/3

What is the first derivative of 1/3 x^6(x^2-3) ?

The first derivative of 1/3 x^6(x^2-3) is (8x^7-18x^5)/3

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\frac{d}{d x} (\frac{1}{3} x^{6} (x^{2} - 3))

\frac{8 x ^{7} - 18 x ^{5}}{3}

Solution steps

\frac{d}{d x} (\frac{1}{3} x^{6} (x^{2} - 3))

Take the constant out:

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

= \frac{1}{3} \frac{d}{d x} (x^{6} (x^{2} - 3))

Apply the Product Rule:

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

f = x^{6}, g = x^{2} - 3

= \frac{1}{3} (\frac{d}{d x} (x^{6}) (x^{2} - 3) + \frac{d}{d x} (x^{2} - 3) x^{6})

\frac{d}{d x} (x^{6}) = 6 x^{5}

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

= \frac{1}{3} (6 x^{5} (x^{2} - 3) + 2 x x^{6})

Simplify

\frac{1}{3} (6 x^{5} (x^{2} - 3) + 2 x x^{6}) : \frac{8 x ^{7} - 18 x ^{5}}{3}

= \frac{8 x ^{7} - 18 x ^{5}}{3}

\frac{d y}{d x} + x y = e^{x} y

\frac{d}{d x} (\frac{( x ^{4} - 4 x + 3 )}{x + 1})

f (x) = x^{2} + 4, at x = 1

f (x) = x^{5} - 3 (x)^{3} + 6

6 x^{5} (x^{4} - 8)

What is the derivative of 1/3 x^6(x^2-3) ?
The derivative of 1/3 x^6(x^2-3) is (8x^7-18x^5)/3
What is the first derivative of 1/3 x^6(x^2-3) ?
The first derivative of 1/3 x^6(x^2-3) is (8x^7-18x^5)/3