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y=(sinx+Cosx)^3求导

使用对数恒等式即可 y=(sinx)^cosx 显然sinx=e^ln(sinx) 所以得到 y=e^[ln(sinx)*cosx] 于是对x求导得到 y'=e^[ln(sinx)*cosx] *[ln(sinx)*cosx]' =(sinx)^cosx * [cosx/sinx *cosx +ln(sinx) *(-sinx)] =(sinx)^cosx * [(cosx)^2/ sinx -sinx *...

(sinxcosx)′ =(sinx) ′cosx+sinx(cosx) ′ =cosxcosx+sinx(-sinx) =(cosx)^2-(sinx)^2 =cos2x 求导法则:f(x)g(x)=f′(x)g(x)+f(x)g′(x)

y'=x'·cosx+x·(cosx)'+(sinx)' =1·cosx+x·(-sinx)+cosx =cosx-xsinx+cosx =2cosx-xsinx 用到的公式: (uv)'=u'v+uv' (cosx)'=-sinx (sinx)'=cosx

y=(sinx)*(cosx)+(cosx)*(sinx) y1=(sinx)^(cosx) lny1=cosxlnsinx y1'/y1=-sinx*lnsinx+cos^2x/sinx y1'=(sinx)^(cosx)*(-sinx*lnsinx+cos^2x/sinx) y2=(cosx)^(sinx) lny2=sinxlncosx y2'/y2=cosxlncosx-sin^2x/cosx y2'=(cosx)^(sinx)*(cosxl...

y=xsin(-x)cosx = -xsinx.cosx y'=-(sinx.cosx + x(cosx)^2 - xsinx.sinx)

解:令t=sinx+cosx,则:因为:sinx+cosx=√2sin(x+π/4),因此:t∈[1,√2]又因为:(sinx+cosx)²=t²=1+2sinxcosx∴sinxcosx=(t²-1)/2则:y=t+(t²-1)/2=(1/2)(t+1)²-1因此:当t=-1时,y有最小值:-1,但t取不到,因此:根...

解由题知 y'=(x^3+3sinx-e^x)' =(x^3)'+(3sinx)'-(e^x)' =3x^2+3cosx-e^x 此题应用的求导公式为 (x^n)'=nx^(n-1) (sinx)'=cosx (e^x)'=e^x 和导数运算法则(af(x)+bg(x)+ch(x))'=af'(x)+bg'(x)+ch'(x)

一阶导数=2xcosx-sinx*x^2 二阶导数=2cosx-2xsinx-cosx*x^2-2xsinx =-4xsinx+2cosx-cosx*x^2 三阶导数=-4sinx-4xcosx-2sinx+sinx*x^2-2xcosx =-6sinx-6xcosx+sinx*x^2

解:y=(1+3^2)^1/2sin(x+p) tanp=3/1=3 p=arctan3 -pai/2

y'=x^4sinx+cosx y=∫(x^4sinx+cosx)dx =∫x^4sinxdx+∫cosxdx =-∫x^4dcosx+sinx =-x^4cosx+∫cosxdx^4+sinx =-x^4cosx+4∫cosx*x^3dx+sinx =-x^4cosx+4∫x^3dsinx+sinx =-x^4cosx+4x^3sinx-4∫sinxdx^3+sinx =-x^4cosx+4x^3sinx-12∫x^2sinxdx+sinx =-x...

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