derivative of e^{x+y}-1

解答

\frac{d}{d x} (e^{x + y} - 1)

e^{x + y} (\frac{d y}{d x} + 1)

求解步骤

\frac{d}{d x} (e^{x + y} - 1)

使用微分加减法定则:

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

= \frac{d}{d x} (e^{x + y}) - \frac{d}{d x} (1)

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

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

= e^{x + y} (1 + \frac{d y}{d x}) - 0

化简

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

\int \frac{sin ^{4} ( x )}{cos ^{6} ( x )} d x

t an g e n t f (x) = 7 x, at x = 4

n = 5 \sum \infty \frac{n ^{3}}{2 ^{n}}

f^{^{''}} (x) = 2

x \to - \infty lim (x + \frac{1}{x})