神马作文网若f(x^2)求导=1 x(其中X>0),且f(1)=2,则f(x)等于多少
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F（X)的导数=2X/(1+X^2)
f(x)为n+1阶多项式,所以n+1阶求导后只会剩下x的n+1次方的导数,为n+1的阶乘
f`(x)=(x+1)`(2x+1)(3x+1)(4x+1)+(x+1)(2x+1)`(3x+1)(4x+1)+(x+1)(2x+1)(3x+1)`(4x+1)+(x+1)(2x+1)(3x+1)(4
有点慢你能等等吗?再答：
f’(x)=[x-x^3)’(1+2x^2+x^4)-(x-x^3)(1+2x^2+x^4)’]/(1+2x^2+x^4)^2=[(1-3x^2)(1+2x^2+x^4)-(x-x^3)(4x+4^3
自己抄把 猴岛外宣Iim 温眸-为你解答
f(x)=(x-1)(x-2)(x-3)f'(x)=(x-2)(x-3)+(x-1)(x-3)+(x-1)(x-2)=x^2-5x+6+x^2-4x+3+x^2-3x+2=3x^2-12x+11
f(x)=2x/x^2+1是f(x)=2x/（x^2+1）的意思噻?f′(x)=[2（x^2+1）-2x*2x]/（x^2+1）^2=(2-2x^2)/（x^2+1）^2u/v求导为：(u′v-uv′
z=y/f(x^2-y^2)ðz/ðx=y(-2xf'/f^2)ðz/ðy=1/f+y(2yf'/f^2)(1/x)*(ðz/ðx)=-2yf'/f^2
f(x)=ln√(x²+1)f'(x)=[1/√(x²+1)]*(√(x²+1))'=[1/√(x²+1)]*[1/2√(x²+1)]*(x²
y=f(x)=x^2xlny=2xlnx对x求导(1/y)*y'=2*lnx+2x*1/x=2lnx+2所以y'=y*(2lnx+2)所以f'(x)=x^2x(2lnx+2)
g'(x)=1/2/√{1+[sinf(x)]^2}*2sinf(x)cosf(x)f'(x)=sinf(x)cosf(x)f'(x)/√{1+[sinf(x)]^2}
利用对数求导法：（可以先取绝对值,不取结果也一样）lnf(x)=lnx+ln(x+1)+...+ln(x+n)1/f(x)*f'(x)=1/x+1/(x+1)+1/(x+2)+...+1/(x+n)所
f(x)=ln(x+√1+x^2)f'(x)=1/(x+√(1+x^2)*(x+√1+x^2)'=1/(x+√(1+x^2)*(1+(√1+x^2)'=1/(x+√(1+x^2)*(1+1/2*√(x
f'(x)=5(2x+1)^4*(2x+1)'=102(2x+1)^4再问：答案错啦再答：哦，多打了一个2f'(x)=5(2x+1)^4*(2x+1)'=10(2x+1)^4
f`(x)=-2^(-x)ln2再问：过程再答：a^x导=a^xlna这是个复合导f`(x)=2^(-x)ln2*(-x)`=-2^(-x)ln2
f(x)=(2x+1)(x-1)=2x^2-2x+x-1=2x^2-x-1f'(x)=4x-1
f(x)=(xlnx)^(-1)所以f'(x)=-1*(xlnx)^(-2)*(xlnx)'(xlnx)'=x'lnx+x*(lnx)'=lnx+x*1/x=lnx+1(xlnx)^(-2)=1/(x
一般的[f(x)/g(x)]'=[f'(x)g(x)-f(x)g'(x)]/[g^2(x)]所以对本题目f'(x)=[e^x*(x-1)-e^x*1]/(x-1)^2=e^x*(x-2)/(x-1)^