derivative 4^x+1

解答

导数

4^{x} + 1

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

求解步骤

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

使用微分加减法定则:

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

= \frac{d}{d x} (4^{x}) + \frac{d}{d x} (1)

\frac{d}{d x} (4^{x}) = ln (2) \cdot 2^{2 x + 1}

\frac{d}{d x} (1) = 0

= ln (2) \cdot 2^{2 x + 1} + 0

化简

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

\frac{d}{d a} (\frac{a ^{2} - 2 a + a}{9})

d er i v a t i v e y^{2} x + 5 x y^{3} + 10 y^{2} + 8 x y

\frac{d}{d a} (\frac{( cos ( a ) )}{( 1 - sin ( a ) )})

d er i v a t i v e y = \frac{1}{( x ^{2} + 4 )}

\frac{d}{d x} (x^{3} sin (y^{3}))