What is the derivative of 2(9e^{3x}x+6e^{3x}) ?

The derivative of 2(9e^{3x}x+6e^{3x}) is 2(27e^{3x}x+27e^{3x})

What is the first derivative of 2(9e^{3x}x+6e^{3x}) ?

The first derivative of 2(9e^{3x}x+6e^{3x}) is 2(27e^{3x}x+27e^{3x})

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\frac{d}{d x} (2 (9 e^{3 x} x + 6 e^{3 x}))

2 (27 e^{3 x} x + 27 e^{3 x})

Solution steps

\frac{d}{d x} (2 (9 e^{3 x} x + 6 e^{3 x}))

Take the constant out:

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

= 2 \frac{d}{d x} (9 e^{3 x} x + 6 e^{3 x})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= 2 (\frac{d}{d x} (9 e^{3 x} x) + \frac{d}{d x} (6 e^{3 x}))

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

\frac{d}{d x} (6 e^{3 x}) = 18 e^{3 x}

= 2 (9 (3 e^{3 x} x + e^{3 x}) + 18 e^{3 x})

Expand

9 (3 e^{3 x} x + e^{3 x}) + 18 e^{3 x} : 27 e^{3 x} x + 27 e^{3 x}

= 2 (27 e^{3 x} x + 27 e^{3 x})

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

\frac{d}{d x} (x \cdot ln (x) - x)

y = 3 x + 2 cos (x)

y = x^{4} - 3 x^{2} + 8, at x = 1

\int (\frac{1}{x ^{5}} - x^{5} - \frac{1}{9}) d x

What is the derivative of 2(9e^{3x}x+6e^{3x}) ?
The derivative of 2(9e^{3x}x+6e^{3x}) is 2(27e^{3x}x+27e^{3x})
What is the first derivative of 2(9e^{3x}x+6e^{3x}) ?
The first derivative of 2(9e^{3x}x+6e^{3x}) is 2(27e^{3x}x+27e^{3x})