derivative of 6e^xt+xt

解答

\frac{d}{d x} (6 e^{x} t + x t)

6 t e^{x} + t

求解步骤

\frac{d}{d x} (6 e^{x} t + x t)

将

t

视为常数

使用微分加减法定则:

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

= \frac{d}{d x} (6 e^{x} t) + \frac{d}{d x} (x t)

\frac{d}{d x} (6 e^{x} t) = 6 t e^{x}

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

= 6 t e^{x} + t

a re a 7 x, 2 x^{2}

\frac{d}{d x} (ln (f (x)) (x))

y^{^{'}} = \frac{sec ( y )}{( x - 1 ) ^{2}}

\frac{d y}{d x} - \frac{2 y}{x} + x^{2} y^{2} = 0

t an g e n t f (x) = 19 x - x^{2}, at x = 9