derivative f(x)=2^xlog_{2}(x)

解答

导数

f (x) = 2^{x} lo g_{2} (x)

2^{x} ln (x) + \frac{2 ^{x}}{ln ( 2 ) x}

求解步骤

\frac{d}{d x} (2^{x} lo g_{2} (x))

使用微分乘法定则:

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

f = 2^{x}, g = lo g_{2} (x)

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

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

\frac{d}{d x} (lo g_{2} (x)) = \frac{1}{x ln ( 2 )}

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

化简

2^{x} ln (2) lo g_{2} (x) + \frac{1}{x ln ( 2 )} \cdot 2^{x} : 2^{x} ln (x) + \frac{2 ^{x}}{ln ( 2 ) x}

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