幾題連鎖率的題目 請數學高手幫解答~

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親愛的[阿棋]您好，[hy]很高興有為您服務的機會，希望能解決您的疑問
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◎第1題參考解法◎
f(x)=[ (x-1)/(x+2) ]^(-1/3)
f'(x)=(-1/3)[ (x-1)/(x+2) ]^(-1/3-1)[ (x-1)/(x+2) ]'
=(-1/3)[ (x-1)/(x+2) ]^(-4/3){[(x-1)'(x+2)-(x+2)'(x-1)]/(x+2)^2}
=(-1/3)[ (x-1)/(x+2) ]^(-4/3){[1(x+2)-(x-1)]/(x+2)^2}
=(-1/3)[ (x-1)/(x+2) ]^(-4/3)[1/(x+2)^2]
=(-1/3)[ (x-1)/(x+2) ]^(-4/3)(x+2)^(-2)------Ans
◎第2題參考解法◎
g(t)=(t^4+1)^(-4) -t^2*√(t^2+1)
g'(t)=
(-4)(t^4+1)^(-4-1)(t^4+1)'-{(t^2)'*(t^2+1)^(1/2)+t^2*[(t^2+1)^(1/2)]'}
g'(t)=
(-4)(t^4+1)^(-5)(4t^3)-{2t(t^2+1)^(1/2)+t^2*(1/2)(t^2+1)^(1/2-1)*(t^2+1)'}
g'(t)=-16t^3(t^4+1)^(-5)-2t(t^2+1)^(1/2)-t^2*(1/2)(t^2+1)^(-1/2)*2t
g'(t)=-16t^3(t^4+1)^(-5)-2t(t^2+1)^(1/2)-t^3(t^2+1)^(-1/2)----Ans
◎第3題參考解法◎
y(x)=[(1+√x)/(1-√x) ]^(-2)
y(x)={[1+x^(1/2)]/[1-x^(1/2)]}^(-2)
y'(x)=(-2){[1+x^(1/2)]/[1-x^(1/2)]}^(-2-1){[1+x^(1/2)]/[1-x^(1/2)]}'
y'(x)=
(-2){[1+√x]/[1-√x]}^(-3){[1+x^(1/2)]'[1-x^(1/2)]-[1-x^(1/2)]'[1+x^(1/2)]}/{[1-x^(1/2)]}^2
y'(x)=
(-2){[1+√x]/[1-√x]}^(-3){[(1/2)x^(1/2-1)[1-x^(1/2)]-[-(1/2)x^(1/2-1)][1+x^(1/2)]}/{[1-x^(1/2)]}^2
y'(x)=
(-2){[1+√x]/[1-√x]}^(-3){[(1/2)x^(-1/2)[1-x^(1/2)]+(1/2)x^(-1/2)[1+x^(1/2)]}/{[1-x^(1/2)]}^2
y'(x)=
(-2)[(1+√x)/(1-√x) ]^(-3)(1/2){[(1-√x) /√x]+[(1+√x)/√x]}/[1-√x]^2
y'(x)=-[(1+√x)/(1-√x) ]^(-3)[2/√x]/[1-√x]^2
y'(x)=-[(1+√x)^(-3)]/[(1-√x)]^(-3)[2/√x]/[1-√x]^2
y'(x)=-[2/√x][(1+√x)^(-3)]/[(1-√x)]^(-3+2)
y'(x)=-2[(1+√x)^(-1)]/√x*[(1-√x)]^(-1)
y'(x)=[-2(1-√x)^(-1)*x^(-1/2)]/[(1+√x)^(3)]------Ans



備註:
(a)連鎖律:df(t)/dx=[df(t)/dt]*[dt/dx]
(b)d[f(x)/g(x)]=[f'(x)*g(x)-g,(x)*f(x)]/[g(x)^2]
(c)√x=x^(1/2)
(d)x^a*x^b=x^(a+b)
(e)x^(-a)=1/x^a
(f)dx^n/dx=n*x^(n-1)
以上答案僅供參考，希望對您有幫助！ 祝順利解決問題!!
若有錯誤，歡迎各位提問或寄信 ，謝謝喔！
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by hy




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1.f(X)=( (X-1)/(X+2) )的-1/3平方 ....看不懂你要表達ㄉ是不是
f(x)=[(x-1)/(x+2)]^-2/3


g(t)=(t^4+1)^-4 -t^2√(t^2+1)=f(t)-h(t)
f(t)=(t^4+1)^-4 f'(t)=-4*[(t^4+1)^-5]*4t^3=-16t^3/(t^4+1)^5
h(t)=t^2*√(t^2+1) h'(t)=(2t)*√(t^2+1)+(t^2)*(1/2)/√(t^2+1)
so g'(t)=f'(t)-g'(t)