What is the derivative of (1+x+x^2)^{99} ?

The derivative of (1+x+x^2)^{99} is 99(1+x+x^2)^{98}(2x+1)

What is the first derivative of (1+x+x^2)^{99} ?

The first derivative of (1+x+x^2)^{99} is 99(1+x+x^2)^{98}(2x+1)

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derivative of

(1 + x + x^{2})^{99}

99 (1 + x + x^{2})^{98} (2 x + 1)

Solution steps

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

Apply the chain rule:

99 (1 + x + x^{2})^{98} \frac{d}{d x} (1 + x + x^{2})

= 99 (1 + x + x^{2})^{98} \frac{d}{d x} (1 + x + x^{2})

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

= 99 (1 + x + x^{2})^{98} (2 x + 1)

\int e^{5 x^{2} + 10 x + 10} (x + 1) d x

\frac{d}{d x} (sin (\frac{x}{8}))

x \to 2 π - lim (e^{\frac{1}{x}} - x)

x \to - \frac{π}{2} lim (csc (\frac{x}{2}))

ln (1 + 2 x)

What is the derivative of (1+x+x^2)^{99} ?
The derivative of (1+x+x^2)^{99} is 99(1+x+x^2)^{98}(2x+1)
What is the first derivative of (1+x+x^2)^{99} ?
The first derivative of (1+x+x^2)^{99} is 99(1+x+x^2)^{98}(2x+1)