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10sinx

∫sinx/x dx =∫sinx dx ∫e^(-xy) dy (两个积分限都一样,0到∞) =∫dy ∫sinx e^(-yx) dx =...(用两次分部积分) =∫dy/(1+y^2) =ArcTan y |^∞_0 =π/2

y=1时 y=x/10中 x=10 y=sinx 的最大值为13*3.145

解:设y=10^sinx lny = sinx * ln10 y'/y = ln10 * cosx y' = ln10 * cosx * y = ln10 * cosx * (10^sinx) 答:10^sinx 的导数是 ln10 * cosx * (10^sinx)。

可用降阶公式 验算结果也正确

double sinx(double x){ double result=x,temp=x; double den=x,fac=1; int n=1,sign=1; while((temp>1e-5)||(temp0?result+temp:result-temp; } return result;} double cosx(double x){ x=1.57079-x; return sinx(x);} main(){ double a,b,c; ...

泰勒展开是这个:sinx=x-x^3/3!+x^5/5!-.. 下面给出算20项的程序。 #include"math.h" #include"stdio.h" void main() { double x=0,y=0,z=1,s=1,mynum=0; int i=1 ,j=0, k=1; scanf("x=%f",&x); for(i=1;i

Dim SinX, SinX2 As Double Dim n As Integer Dim dlt As Double Dim X As Single Dim PI As Double = 3.14159265358979 Dim signV As Integer Dim pMup As Double X = InputBox("输入数字") X = ((PI + X) Mod 2 * PI) - PI ' 保证 X 位于 -π~...

先降幂 I = ∫x(sinx)^10dx = (1/32)∫x(1-cos2x)^5dx = (1/32)∫x[1-5cos2x+10(cos2x)^2-10(cos2x)^3 +5(cos2x)^4-(cos2x)^5]dx = (1/32)∫x[1-5cos2x+5(1+cos4x)-10(cos2x)^3 +(5/4)(1+cos4x)^2-(cos2x)^5]dx = (1/32)∫x[6-5cos2x+5cos4x-10(cos2x...

y=1时 y=x/10中 x=10 y=sinx 的最大值为13*3.145

可用四元均值不等式: y=10sinxcosx+10sinx =10sinx(1+cosx) =20sin(x/2)·cos(x/2)·2cos²(x/2) =40sin(x/2)cos³(x/2) ∴y²=(1600/3)·3sin²(x/2)·cos²(x/2)·cos²(x/2)·cos²(x/2) ≤(1600/3)·[(3sin²(x/2)+3...

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