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

\frac{\partial}{\partial u} ((u^{2} + v^{2})^{\frac{3}{2}})

x \to 4 - lim (\frac{1}{x - 4})

\int (x^{2} - 1)^{- 1} d x

x^{2} + 2 y^{2} = 2

x^{2} + x - 2, [- 3, 4]

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)