y=arctan(1-x的平方)是复合函数吗
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两边取正切y=tan(x+1)
令F(x)=2x/(1+x^2)=2/(x+(1/x))而由重要不等式可知:x+1/x≥2或x+1/x≤-2所以易得F(x)=2x/(1+x^2)的值域为[-1,1]所以y=arctan(2x/1+x
y=arctanx/(1+x²)那么y'=1/[1+x²/(1+x²)²]*[x/(1+x²)]'=(1+x²)²/[(1+x
[(2x²-3)³]=2(2x²-3)²*4x=8x(2x²-3)²(arctanx)'=1/(1+x²)所以y'=[8x(2x&
y=arctan(1-x)1-x=tany对x求导-1=y'sec²y所以y'=-1/sec²y=-cos²y=-cos²[arctan(1-x)]y'=-co
两边同时求导根据链式法则1/2(x²+y²)’/(x²+y²)=(x/y)'/[1+(x/y)²]1/2(2x+2yy')/(x²+y
y'=1/[1+(1/x)^2]*(1/x)'=x^2/(1+x^2)*(-1/x^2)=-1/(1+x^2)
y=4arctanxy'=4/(1+x^2)所以y'(1)=4/(1+1^2)=2
y'=1/[1+(x^2+1)^2]×(x^2+1)'=2x/(x^4+2x^2+2)再问:
须知(e^x)'=e^x,(arctanx)'=1/(1+x²)y=e^arctan(1/x)y'=e^arctan(1/x)·1/[1+(1/x)²]·(-1/x²)=
tany=1y可以有无穷多个值但是前面几步(arctanx∈(-π/2,-π/4)∪(-π/4,π/2),arctan(1-x)/(1+x)∈(-π/2,-π/4)∪(-π/4,π/2))限制y大于-
令m=arctanx,n=arctan(1-x)/(1+x)那么x=tanm,(1-x)/(1+x)=tann,y=m+n那么tany=tan(m+n)=(tanm+tann)/(1-tanntanm
此题复合求导dy=d[arctan(1-x/1+x)]=[1/(1+(1-x/1+x)^2)]·(1-x/1+x)';注:(arctanx)'=1/(1+x^2)=-(1/(x^2+1))
y=arctanx+1\x-1y'=1/[1+(x+1\x-1)^2]*(x+1\x-1)'=1/[1+(x+1\x-1)^2]*(-2)/(x-1)^2=-1/(1+x^2)
此题是这样的吧:函数y=arctan[(1+x)/(1-x)]?若是这样,y′=1/[1+(1+x)²/(1-x)²][(1-x)+(1+x)]/(1-x)²=2/[(1