derivative of 3^xlog_{5}(x)

解答

\frac{d}{d x} (3^{x} lo g_{5} (x))

\frac{3 ^{x} + ln ( 3 ) \cdot 3 ^{x} x ln ( x )}{ln ( 5 ) x}

求解步骤

\frac{d}{d x} (3^{x} lo g_{5} (x))

使用微分乘法定则:

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

f = 3^{x}, g = lo g_{5} (x)

= \frac{d}{d x} (3^{x}) lo g_{5} (x) + \frac{d}{d x} (lo g_{5} (x)) 3^{x}

\frac{d}{d x} (3^{x}) = 3^{x} ln (3)

\frac{d}{d x} (lo g_{5} (x)) = \frac{1}{x ln ( 5 )}

= 3^{x} ln (3) lo g_{5} (x) + \frac{1}{x ln ( 5 )} \cdot 3^{x}

化简

3^{x} ln (3) lo g_{5} (x) + \frac{1}{x ln ( 5 )} \cdot 3^{x} : \frac{3 ^{x} + ln ( 3 ) \cdot 3 ^{x} x ln ( x )}{ln ( 5 ) x}

= \frac{3 ^{x} + ln ( 3 ) \cdot 3 ^{x} x ln ( x )}{ln ( 5 ) x}

\int_{0}^{x} \frac{1}{1 - x} d x

\frac{d}{d θ} (\frac{1}{θ})

\frac{\partial}{\partial x} (6 e^{x^{3} y})

t an g e n t f (x) = 3 x^{3} - x^{2} + 8, (2, 28)

\frac{d}{d x} (8 - x^{2})