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

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

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

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

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derivative of

f (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)

f (x) = \frac{1}{5 + 3 x}, (2, \frac{1}{11})

\int_{1}^{2} 8 r^{2} ln (r) d r

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f (x) = \frac{x ^{3}}{3} - \frac{3}{x ^{3}}, at x = - 1

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