数学函数图像为您作e^sqrt(x)的函数图像。
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{\sqrt{x}} $$dx 令$$ \sqrt{x}=M $$ 则$$ x=u^{2} $$ $$ dx=2udu $$ $$ 原式= \int _{2}ue^{u}du \\ =2 \int _{m}de^{\mu} \\ =2ue^{u}-2 \int e^{\mu}du \\ =2 \mu e^{\mu}-2e^{\mu}+C \\ =2 \sqrt{x}e^{\sqrt{x}}-2e^{\sqrt{x}}+c $...
x ਦੇ ਬਾਬਤ sqrt(e^sqrtx), x>0 ਦਾ ਡੈਰੀਵੇਟਿਵ ਪਤਾ ਕਰੋ।
f$$ e ^ { \sqrt { x } } d x $$; 相关知识点: 试题来源: 解析 解$$ \sqrt{x}=t $$.则$$ x=t^{2} $$ 则$$ d_{x}=2+dt $$ i $$ e^{t}dt=2e^{t}(t-1)+c \\ =2e^{\sqrt{x}}(\sqrt{x}-1)+c $$ ((为弟数) ...
For the exponential function, dxd(ef(x))=ef(x)f′(x). Here, ef(x)=e3x, so f(x)=3x=3x1/2. So then we must have dxd(e3x)=e3x(23x−1/2)=2x3e3x ... x2=ex and x=sqrt(ex). When both are the same equations, why do we get a different value of x ? https://www....
此时我们利用 \arctan x-l_n(x) 和u_n(x)-\arctan x 的导数来构造积分。定义 \color{red}{s_n(x)=\left(\arctan x-l_n(x)\right)'=\dfrac{1}{1+x^2}-l_n'(x)} 为下界误差函数,以及 \color{blue}{t_n(x)=\left(u_n(x)-\arctan x\right)'=l_n'(x)-\dfrac{1}{1+x^2...
解 我们首先想到要去掉根号,为此,令$$ \sqrt { x } = t , x = t ^ { 2 } $$,则 $$ \int e ^ { \sqrt { x } } d x = \int e ^ { t } \cdot 2 t d t = 2 \int t e ^ { t } d t , $$ 此时再用分部积分法.利用例2的结果,并用$$ t = \sqrt { x } $$代回...
一元二次方程ax^2+bx+c=0(a\neq 0)的万能公式：其根可以表示为：x=\frac{-b\pm \sqrt{b^2...
$$ \sqrt{e^{- \sqrt{x}}dx} $$ 解:&今$$ - \sqrt{x}=t x=t^{2} $$ $$ 原式= \int e^{t}d(t^{2})= \sqrt{e}^{t}\cdot 2t \cdot dt $$ $$ =2 \int _{t}\cdot d(e^{t})=2(t \cdot e^{t}- \int e^{t}\cdot alt) $$ $$ =2(te^{t}-e^{t})=2e^...