derivative of a+b^1x^1+b^2x^2

解

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

2 b^{2} x + b

解答ステップ

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

簡素化

a + b^{1} x^{1} + b^{2} x^{2} : a + b x + b^{2} x^{2}

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

和/差の法則を適用:

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

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

\frac{d a}{d x} = 0

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

\frac{d}{d x} (b^{2} x^{2}) = 2 b^{2} x

= 0 + b + 2 b^{2} x

簡素化

= 2 b^{2} x + b