What is the derivative of 6xsqrt(x^3+3x) ?

The derivative of 6xsqrt(x^3+3x) is (3(5x^3+9x))/(sqrt(x^3+3x))

What is the first derivative of 6xsqrt(x^3+3x) ?

The first derivative of 6xsqrt(x^3+3x) is (3(5x^3+9x))/(sqrt(x^3+3x))

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\frac{d}{d x} (6 x x^{3} + 3 x)

\frac{3 ( 5 x ^{3} + 9 x )}{x ^{3} + 3 x}

Solution steps

\frac{d}{d x} (6 x x^{3} + 3 x)

Take the constant out:

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

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

Apply the Product Rule:

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

f = x, g = x^{3} + 3 x

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

\frac{d x}{d x} = 1

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

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

Simplify

6 (1 \cdot x^{3} + 3 x + \frac{3 x ^{2} + 3}{2 x ^{3} + 3 x} x) : \frac{3 ( 5 x ^{3} + 9 x )}{x ^{3} + 3 x}

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

\frac{\partial}{\partial x} (x h)

y^{^{'}} + 6 y = t + e^{- 2 t}

\frac{d}{d x} (\frac{5 x}{1 + 2 x ^{2}})

x \to 0 + lim (\frac{sin ^{2} ( x )}{x})

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

What is the derivative of 6xsqrt(x^3+3x) ?
The derivative of 6xsqrt(x^3+3x) is (3(5x^3+9x))/(sqrt(x^3+3x))
What is the first derivative of 6xsqrt(x^3+3x) ?
The first derivative of 6xsqrt(x^3+3x) is (3(5x^3+9x))/(sqrt(x^3+3x))