What is the derivative of 1.1x+6.3x^3 ?

The derivative of 1.1x+6.3x^3 is 1.1+18.9x^2

What is the first derivative of 1.1x+6.3x^3 ?

The first derivative of 1.1x+6.3x^3 is 1.1+18.9x^2

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\frac{d}{d x} (1.1 x + 6.3 x^{3})

1.1 + 18.9 x^{2}

Solution steps

\frac{d}{d x} (1.1 x + 6.3 x^{3})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (1.1 x) + \frac{d}{d x} (6.3 x^{3})

\frac{d}{d x} (1.1 x) = 1.1

\frac{d}{d x} (6.3 x^{3}) = 18.9 x^{2}

= 1.1 + 18.9 x^{2}

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

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

y = \frac{2}{x - 4}, (6, 1)

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

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

What is the derivative of 1.1x+6.3x^3 ?
The derivative of 1.1x+6.3x^3 is 1.1+18.9x^2
What is the first derivative of 1.1x+6.3x^3 ?
The first derivative of 1.1x+6.3x^3 is 1.1+18.9x^2