求导 y=x的平方sinx分之一 y=In根号x+根号Inx y=e的xInx次方 求导:y=ln(x+根号下a方加x方)

y=\u5927\u6839\u53f7\u4e0bxsinx\u5c0f\u6839\u53f71-e^x \u7528\u5bf9\u6570\u6c42\u5bfc\u6cd5\u6c42\u51fd\u6570\u7684\u5bfc\u6570

\u3000\u3000\u5bf9\u6570\u6c42\u5bfc\u6cd5\u6559\u6750\u4e0a\u6709\u4f8b\u9898\u7684\uff0c\u4f9d\u6837\u753b\u846b\u82a6\u5373\u53ef\uff1a\u53d6\u5bf9\u6570\uff0c\u5f97
\u3000\u3000\u3000lny = (1/2)lnx+(1/2)lnsinx+(1/4)ln(1-e^x)\uff0c
\u6c42\u5bfc\uff0c\u5f97
\u3000\u3000\u3000y'/y = (1/2)(1/x)+(1/2)tanx+(1/4)[(-e^x)/(1-e^x)]\uff0c
\u6240\u4ee5\uff0c
\u3000\u3000\u3000y' = y*{(1/2)(1/x)+(1/2)tanx+(1/4)[(-e^x)/(1-e^x)]}
\u3000\u3000\u3000\u3000= \u2026\u2026

y\uff07\uff1d1\uff0f\uff08x\uff0b\u6839\u53f7\u4e0bx\u5e73\u65b9\uff0ba\u5e73\u65b9\uff09\uff0a\uff081\uff0bx\uff0f\u6839\u53f7\u4e0bx\u5e73\u65b9\uff0ba\u5e73\u65b9\uff09\uff1d1\uff0f\u6839\u53f7\u4e0bx\u5e73\u65b9\uff0ba\u5e73\u65b9
y\uff1dln\uff08x\uff0b\u221a\uff08x²\uff0ba²\uff09\uff09
y\u2032\uff1d\uff081\uff0bx\uff0f\u221a\uff08x²\uff0ba²\uff09\uff09\uff0f\uff08x\uff0b\u221a\uff08x²\uff0ba²\uff09\uff09\uff1d1\uff0f\u221a\uff08x²\uff0ba²\uff09
y\u2033\uff1d\uff0dx\uff0f\u221a\uff08x²\uff0ba²\uff09³
\u8fd9\u662f\u53cd\u53cc\u66f2\u6b63\u5f26\u51fd\u6570\u6c42\u5bfc\uff0c
y\uff07\uff1d\uff3b1\uff0b\uff081\uff0f2\uff09\uff0a2x\uff0f\u221a\uff08a\uff3e2\uff0bx\uff3e2\uff09\uff3d\uff0f\uff3bx\uff0b\u221a\uff08a\uff3e2\uff0bx\uff3e2\uff09\uff3d
\uff1d\uff3bx\uff0b\u221a\uff08a\uff3e2\uff0bx\uff3e2\uff09\uff3d\uff0f\u221a\uff08a\uff3e2\uff0bx\uff3e2\uff09\uff0f\uff3bx\uff0b\u221a\uff08a\uff3e2\uff0bx\uff3e2\uff09\uff3d
\uff1d1\uff0f\u221a\uff08a\uff3e2\uff0bx\uff3e2\uff09\u3002
y\uff07\uff1d1\uff0f\uff3bx\uff0b\u6839\u53f7\uff08x\uff3e2\uff0ba\uff3e2\uff09\uff3d\uff0a\uff3bx\uff0b\u6839\u53f7\uff08x\uff3e2\uff0ba\uff3e2\uff09\uff3d\uff07
\uff1d1\uff0f\uff3bx\uff0b\u6839\u53f7\uff08x\uff3e2\uff0ba\uff3e2\uff09\uff3d\uff0a\uff3b1\uff0bx\uff0f\u6839\u53f7\uff08x\uff3e2\uff0ba\uff3e2\uff09\uff3d
\uff1d1\uff0f\u6839\u53f7\uff08x\uff3e2\uff0ba\uff3e2\uff09\u3002

\u6269\u5c55\u8d44\u6599
\u57fa\u672c\u521d\u7b49\u51fd\u6570\u6c42\u5bfc\u516c\u5f0f\uff1a
\uff08C\uff09\uff07\uff1d0\uff0c
\uff08x\uff3ea\uff09\uff07\uff1dax\uff3e\uff08a\uff0d1\uff09\uff0c
\u3000\u3000(a^x)'=(a^x)lna\uff0ca>0\uff0ca\u22601;(e^x)'=e^x
\u3000\u3000[logx]'=1/[xlna]\uff0ca>0\uff0ca\u22601;(lnx)'=1/x

\uff08sinx\uff09\uff07\uff1dcosx
\uff08cosx\uff09\uff07\uff1d\uff0dsinx
\uff08tanx\uff09\uff07\uff1d\uff08secx\uff09\uff3e2
\uff08cotx\uff09\uff07\uff1d\uff0d\uff08cscx\uff09\uff3e2
\uff08arcsinx\uff09\uff07\uff1d1\uff0f\u221a\uff081\uff0dx\uff3e2\uff09
\uff08arccosx\uff09\uff07\uff1d\uff0d1\uff0f\u221a\uff081\uff0dx\uff3e2\uff09
\uff08arctanx\uff09\uff07\uff1d1\uff0f\uff081\uff0bx\uff3e2\uff09
\uff08arccotx\uff09\uff07\uff1d\uff0d1\uff0f\uff081\uff0bx\uff3e2\uff09
\u2461\u56db\u5219\u8fd0\u7b97\u516c\u5f0f
\uff08u\uff0bv\uff09\uff07\uff1du\uff07\uff0bv\uff07
\uff08u\uff0dv\uff09\uff07\uff1du\uff07\uff0dv\uff07
\uff08uv\uff09\uff07\uff1du\uff07v\uff0buv\uff07
\uff08u\uff0fv\uff09\uff07\uff1d\uff08u\uff07v\uff0duv\uff07\uff09\uff0fv\uff3e2
\u2462\u590d\u5408\u51fd\u6570\u6c42\u5bfc\u6cd5\u5219\u516c\u5f0f
y\uff1df\uff08t\uff09\uff0ct\uff1dg\uff08x\uff09\uff0cdy\uff0fdx\uff1df\uff07\uff08t\uff09\uff0ag\uff07\uff08x\uff09
\u2463\u53c2\u6570\u65b9\u7a0b\u786e\u5b9a\u51fd\u6570\u6c42\u5bfc\u516c\u5f0f
x\uff1df\uff08t\uff09\uff0cy\uff1dg\uff08t\uff09\uff0cdy\uff0fdx\uff1dg\uff07\uff08t\uff09\uff0ff\uff07\uff08t\uff09
\u2464\u53cd\u51fd\u6570\u6c42\u5bfc\u516c\u5f0f
y\uff1df\uff08x\uff09\u4e0ex\uff1dg\uff08y\uff09\u4e92\u4e3a\u53cd\u51fd\u6570\uff0c\u5219f\uff07\uff08x\uff09\uff0ag\uff07\uff08y\uff09\uff1d1\u3002

y=x的平方sinx分之一
y'=(x的平方sinx分之一)'=2x/sinx+x^2(1/sinx)'=2x/sinx-x^2(1/sinx^2)*cosx=2x/sinx-x^2cosx/sinx^2
y=In根号x+根号Inx
y'=1/(2根号x)*1/根号x+1/(2根号lnx)*1/x=1/(2x)+1/(2x根号lnx)=(根号lnx+1)/(2x根号lnx)
y=e的xInx次方=e^x+x
y'=e^x+1

你好,相乘形式的导数为:前导乘以后,加上前乘以后导(前面的导数和后面相乘,加上前面和后面的导数相乘)
对于sinx分之一的导数,求法就是先把sinx看作一个整体求导,然后再对sinx求导,两项相乘就可以了。
lnx的导数就是1/x,把x改成根号x的话也是一样方法,把根号x看作整体求导,然后对根号x求导,同样相乘就可以了。

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这种题目我觉得答案的表述有点复杂,写不太清楚,所以你可以自己先试一下,有疑问再讨论哈。

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