第四题对数求导法
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如图。
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>> clear
>> syms x
>> f=sqrt(x*sqrt(x*sqrt(1-sin(x))));
>> df1=diff(f,x,1)
df1 =
((x*(1 - sin(x))^(1/2))^(1/2) + (x*((1 - sin(x))^(1/2) - (x*cos(x))/(2*(1 - sin(x))^(1/2))))/(2*(x*(1 - sin(x))^(1/2))^(1/2)))/(2*(x*(x*(1 - sin(x))^(1/2))^(1/2))^(1/2))
>> df2=diff(f,x,2)
df2 =
- ((x*(1 - sin(x))^(1/2))^(1/2) + (x*((1 - sin(x))^(1/2) - (x*cos(x))/(2*(1 - sin(x))^(1/2))))/(2*(x*(1 - sin(x))^(1/2))^(1/2)))^2/(4*(x*(x*(1 - sin(x))^(1/2))^(1/2))^(3/2)) - ((x*((1 - sin(x))^(1/2) - (x*cos(x))/(2*(1 - sin(x))^(1/2)))^2)/(4*(x*(1 - sin(x))^(1/2))^(3/2)) - ((1 - sin(x))^(1/2) - (x*cos(x))/(2*(1 - sin(x))^(1/2)))/(x*(1 - sin(x))^(1/2))^(1/2) + (x*(cos(x)/(1 - sin(x))^(1/2) + (x*cos(x)^2)/(4*(1 - sin(x))^(3/2)) - (x*sin(x))/(2*(1 - sin(x))^(1/2))))/(2*(x*(1 - sin(x))^(1/2))^(1/2)))/(2*(x*(x*(1 - sin(x))^(1/2))^(1/2))^(1/2))
df1是一阶导数，df2是二阶导数。
>> syms x
>> f=sqrt(x*sqrt(x*sqrt(1-sin(x))));
>> df1=diff(f,x,1)
df1 =
((x*(1 - sin(x))^(1/2))^(1/2) + (x*((1 - sin(x))^(1/2) - (x*cos(x))/(2*(1 - sin(x))^(1/2))))/(2*(x*(1 - sin(x))^(1/2))^(1/2)))/(2*(x*(x*(1 - sin(x))^(1/2))^(1/2))^(1/2))
>> df2=diff(f,x,2)
df2 =
- ((x*(1 - sin(x))^(1/2))^(1/2) + (x*((1 - sin(x))^(1/2) - (x*cos(x))/(2*(1 - sin(x))^(1/2))))/(2*(x*(1 - sin(x))^(1/2))^(1/2)))^2/(4*(x*(x*(1 - sin(x))^(1/2))^(1/2))^(3/2)) - ((x*((1 - sin(x))^(1/2) - (x*cos(x))/(2*(1 - sin(x))^(1/2)))^2)/(4*(x*(1 - sin(x))^(1/2))^(3/2)) - ((1 - sin(x))^(1/2) - (x*cos(x))/(2*(1 - sin(x))^(1/2)))/(x*(1 - sin(x))^(1/2))^(1/2) + (x*(cos(x)/(1 - sin(x))^(1/2) + (x*cos(x)^2)/(4*(1 - sin(x))^(3/2)) - (x*sin(x))/(2*(1 - sin(x))^(1/2))))/(2*(x*(1 - sin(x))^(1/2))^(1/2)))/(2*(x*(x*(1 - sin(x))^(1/2))^(1/2))^(1/2))
df1是一阶导数，df2是二阶导数。
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对不起我是真的看不懂
追答
这是MATLAB编程计算出来的结果，绝对正确。
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