derivative of e^{x^2-4}+3x+5

解答

\frac{d}{d x} (e^{x^{2} - 4} + 3 x + 5)

2 e^{x^{2} - 4} x + 3

求解步骤

\frac{d}{d x} (e^{x^{2} - 4} + 3 x + 5)

使用微分加减法定则:

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

= \frac{d}{d x} (e^{x^{2} - 4}) + \frac{d}{d x} (3 x) + \frac{d}{d x} (5)

\frac{d}{d x} (e^{x^{2} - 4}) = e^{x^{2} - 4} \cdot 2 x

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

\frac{d}{d x} (5) = 0

= e^{x^{2} - 4} \cdot 2 x + 3 + 0

化简

= 2 e^{x^{2} - 4} x + 3

x \to \frac{π}{4} lim (cot (x))

\int_{0}^{\frac{π}{2}} tan (x) d x

\int \frac{x ^{2} + 1}{( x - 10 ) ( x - 9 ) ^{2}} d x

t an g e n t f (x) = (8 - 9 x) (8 + 9 x), at x = 9

(6 + x) y^{^{'}} = 9 y