derivative of 2^x(x^2+1)

解答

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

ln (2) \cdot 2^{x} (x^{2} + 1) + 2^{1 + x} x

求解步骤

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

使用微分乘法定则:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = 2^{x}, g = x^{2} + 1

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

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

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

= 2^{x} ln (2) (x^{2} + 1) + 2 x \cdot 2^{x}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2 \cdot 2^{x} = 2^{1 + x}

= ln (2) \cdot 2^{x} (x^{2} + 1) + 2^{1 + x} x

(2 x - 1) d x + (3 y + 5) d y = 0

\int (1 + 4 x)^{\frac{3}{5}} d x

\frac{\partial}{\partial x} (x sin (z))

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\int 20 x^{19} d x