y=sinx在x=0处的切线方程是 曲线y=x+sinx在点（0，0）处的切线方程是______
\u6b63\u5f26\u51fd\u6570y=sinx\u5728x=0\u5904\u5207\u7ebf\u65b9\u7a0b\u662f______\u51fd\u6570\u7684\u5bfc\u6570f\u2032\uff08x\uff09=cosx\uff0c\u5219\u5728x=0\u5904\u5207\u7ebf\u659c\u7387k=f\u2032\uff080\uff09=cos0=1\uff0c\u5219\u5728x=0\u5904\u5207\u7ebf\u65b9\u7a0b\u4e3ay-0=x-0\uff0c\u5373y=x\uff0c\u6545\u7b54\u6848\u4e3a\uff1ay=x
y'=cosx
\u6240\u4ee5\u66f2\u7ebf\u5728\u70b9(0,0)\u5904\u7684\u5207\u7ebf\u7684\u659c\u7387k=coso=1
\u5219\u5207\u7ebf\u65b9\u7a0b\u53ef\u8bbe\u4e3ay=x+b
\u56e0\u8fc7(0,0)\u70b9,\u5f97b=0
\u6240\u4ee5\u5207\u7ebf\u65b9\u7a0b\u4e3ay=x
\u6269\u5c55\u8d44\u6599
\u82e5\u70b9\u5728\u66f2\u7ebf\u4e0a\uff0c\u516c\u5f0f\u4e3ay-f(a)=f'(a)(x-a)\uff1b\u82e5\u70b9\u4e0d\u5728\u66f2\u7ebf\u4e0a\uff0c\u516c\u5f0f\u4e3ay-f(x0)=f'(x0)(x-x0)\u3002\u5207\u7ebf\u65b9\u7a0b\u662f\u7814\u7a76\u5207\u7ebf\u4ee5\u53ca\u5207\u7ebf\u7684\u659c\u7387\u65b9\u7a0b\uff0c\u6d89\u53ca\u51e0\u4f55\u3001\u4ee3\u6570\u3001\u7269\u7406\u5411\u91cf\u3001\u91cf\u5b50\u529b\u5b66\u7b49\u5185\u5bb9\u3002
\u5982\u679c\u67d0\u70b9\u5728\u66f2\u7ebf\u4e0a\uff1a\u8bbe\u66f2\u7ebf\u65b9\u7a0b\u4e3ay=f(x)\uff0c\u66f2\u7ebf\u4e0a\u67d0\u70b9\u4e3a(a\uff0cf(a))\u6c42\u66f2\u7ebf\u65b9\u7a0b\u6c42\u5bfc\uff0c\u5f97\u5230f'(x)\uff0c
\u5c06\u67d0\u70b9\u4ee3\u5165\uff0c\u5f97\u5230f'(a)\uff0c\u6b64\u5373\u4e3a\u8fc7\u70b9(a\uff0cf(a))\u7684\u5207\u7ebf\u659c\u7387\uff0c\u7531\u76f4\u7ebf\u7684\u70b9\u659c\u5f0f\u65b9\u7a0b\uff0c\u5f97\u5230\u5207\u7ebf\u7684\u65b9\u7a0b\u3002y-f(a)=f'(a)(x-a)\u3002
求导得：y'=cosx
切线斜率k=y'|(x=0)=cos0=1
x=0时， y=sin0=0，∴切点（0,0）
所以切线方程为 y=x
y=sinx
求导
y'=cosx x=0时 y'=1 y=sin0=0
所以
切线为 y=x-1
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绛旓細杩欐牱
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