補(bǔ)檔數(shù)學(xué)公式
忍不住了,██!??????
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? ? ? ? ?(sinx)' = cosx
(cosx)' = - sinx
(tanx)'=1/(cosx)^2=(secx)^2=1+(tanx)^2
-(cotx)'=1/(sinx)^2=(cscx)^2=1+(cotx)^2
(secx)'=tanx·secx
(cscx)'=-cotx·cscx
(arcsinx)'=1/(1-x^2)^1/2
(arccosx)'=-1/(1-x^2)^1/2
(arctanx)'=1/(1+x^2)
(arccotx)'=-1/(1+x^2)
(arcsecx)'=1/(|x|(x^2-1)^1/2)
(arccscx)'=-1/(|x|(x^2-1)^1/2)
?、?sinhx)'=coshx
(coshx)'=sinhx
(tanhx)'=1/(coshx)^2=(sechx)^2
(coth)'=-1/(sinhx)^2=-(cschx)^2
(sechx)'=-tanhx·sechx
(cschx)'=-cothx·cschx
(arsinhx)'=1/(x^2+1)^1/2
(arcoshx)'=1/(x^2-1)^1/2
(artanhx)'=1/(x^2-1) (|x|<1)
(arcothx)'=1/(x^2-1) (|x|>1)
(arsechx)'=1/(x(1-x^2)^1/2)
(arcschx)'=1/(x(1+x^2)^1/2)
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∫0dx=c ∫x^udx=(x^(u+1))/(u+1)+c∫1/xdx=ln|x|+c∫a^xdx=(a^x)/lna+c∫e^xdx=e^x+c∫sinxdx=-cosx+c∫cosxdx=sinx+c∫1/(cosx)^2dx=tanx+c∫1/(sinx)^2dx=-cotx+c∫1/√(1-x^2) dx=arcsinx+c∫1/(1+x^2)dx=arctanx+c∫1/(a^2-x^2)dx=(1/2a)ln|(a+x)/(a-x)|+c∫secxdx=ln|secx+tanx|+c ∫1/(a^2+x^2)dx=1/a*arctan(x/a)+c∫1/√(a^2-x^2) dx=(1/a)*arcsin(x/a)+c∫sec^2 x dx=tanx+c;∫shx dx=chx+c;∫chx dx=shx+c;∫thx dx=ln(chx)+c;