What is the derivative of (1+2x+3x^2+4x^3^{100}) ?

The derivative of (1+2x+3x^2+4x^3^{100}) is 100(1+2x+3x^2+4x^3)^{99}(12x^2+6x+2)

What is the first derivative of (1+2x+3x^2+4x^3^{100}) ?

The first derivative of (1+2x+3x^2+4x^3^{100}) is 100(1+2x+3x^2+4x^3)^{99}(12x^2+6x+2)

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\frac{d}{d x} ((1 + 2 x + 3 x^{2} + 4 x^{3})^{100})

100 (1 + 2 x + 3 x^{2} + 4 x^{3})^{99} (12 x^{2} + 6 x + 2)

Solution steps

\frac{d}{d x} ((1 + 2 x + 3 x^{2} + 4 x^{3})^{100})

Apply the chain rule:

100 (1 + 2 x + 3 x^{2} + 4 x^{3})^{99} \frac{d}{d x} (1 + 2 x + 3 x^{2} + 4 x^{3})

= 100 (1 + 2 x + 3 x^{2} + 4 x^{3})^{99} \frac{d}{d x} (1 + 2 x + 3 x^{2} + 4 x^{3})

\frac{d}{d x} (1 + 2 x + 3 x^{2} + 4 x^{3}) = 12 x^{2} + 6 x + 2

= 100 (1 + 2 x + 3 x^{2} + 4 x^{3})^{99} (12 x^{2} + 6 x + 2)

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2 t y^{^{'}} = 4 y

\frac{d P}{d t} = 0.08 P (1 - \frac{P}{1000})

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What is the derivative of (1+2x+3x^2+4x^3^{100}) ?
The derivative of (1+2x+3x^2+4x^3^{100}) is 100(1+2x+3x^2+4x^3)^{99}(12x^2+6x+2)
What is the first derivative of (1+2x+3x^2+4x^3^{100}) ?
The first derivative of (1+2x+3x^2+4x^3^{100}) is 100(1+2x+3x^2+4x^3)^{99}(12x^2+6x+2)