float x=1,y;y=x3 2;则y的值
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x3+y3=100(x+y)(x^2-xy+y^2)=100因x+y=1所以x^2-xy+y^2=100(x+y)^2-3xy=1001-3xy=100xy=-33x^2+y^2=(x+y)^2-2x
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曲线y=x3+x-2求导可得y′=3x2+1设切点为(a,b)则3a2+1=4,解得a=1或a=-1切点为(1,0)或(-1,-4)与直线4x-y-1=0平行且与曲线y=x3+x-2相切的直线方程是:
解题思路:利用导数的知识求解。解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/re
f'(x)=3x²+2k=f'(0)=2f(0)=1所以切线方程为y=2(x-0)+1即2x-y+1=0
由题意,得斜率=3×1平方=3所以切线方程为y-1=3(x-1)即y=3x-2
y=1/x+2/x²+1/x³y'=-1/x²-4/x³-3/x⁴再问:可以解释一下怎么求导么?为什么2/X²导数是-4/X³1
y'=3x2∴y'|x=1=3,而切点的坐标为(1,2)∴曲线y=x3+1在x=1的处的切线方程为y-2=3(x-1),即y=3x-1故选B.
x3次方y-2x2y2+xy3=xy(x²-2xy+y²)=xy(x-y)²=3x3²=27如果本题有什么不明白可以追问,再问:=xy(x2-2xy+y2)=x
(x+y)³=x³+y³+3x²y+3xy².记忆方法:各立方,然后3x方y,3xy方(x+y)³=x³-y³-3x
y′=3x2+1>0∴函数y=x3+x的递增区间是(-∞,+∞),故选C
∵x+y=1∴x3+y3-xy=(x+y)(x2+y2-xy)-xy=x2+y2-2xy=(x-y)2≥0即x+y=1时,x3+y3-xy的值是非负数.
y'=(X^3)'+(xsinx)'=3x^2+(x)'sinx+x(sinx)'=3x^2+sinx+xcosx
x+y=1(x+y)^2=x^2+2xy+y^2=1(x+y)^3=x^3+y^3+3xy(x+y)=1而x^3+y^3=1/3,代入得:3xy=2/3xy=2/9由于x=1-y;故代入xy=2/9;
设f(x)=x3+sinx,则函数的定义域为R∵f(-x)=−x3+sin(−x)=-(x3+sinx)=-f(x)∴函数为奇函数∵f′(x)=13+cosx,∴函数在原点右侧,靠近原点处单调增故选C
答:y=x^3求导:y'(x)=3x^2x=1时:y(1)=1,y'(1)=3切点(1,1),切线斜率k=3切线方程y-1=3(x-1)所以:切线为y=3x-2
∵y=x3+x∴y′=3x2+1.令y′=4⇒x2=1⇒x=±1.把x=1代入y=x3+x得:y=2.所以切线方程为:y-2=4(x-1)⇒4x-y-2=0;把x=-1代入y=x3+x得:y=-2,所
y′=3x2-1≥-1,∴tanα≥-1,∴[0,π2)∪[3π4,π),故答案为[0,π2)∪[3π4,π)
y=x³-6x²+12x-8-x³=-6x²+12x-8=-6(x-1)²-2所以x=1,y最大=-2
x3+3xy-y3=(x-y)(x^2+y^2+xy)+3xy=-x^2-y^2+2xy=-(x-y)^2=-1