隐函數求導法則講解?dⅹ²/dⅹ=2ⅹdy²/dⅹ，令μ=y²→，我來為大家講解一下關于隐函數求導法則講解?跟着小編一起來看一看吧!
dⅹ²/dⅹ=2ⅹ
dy²/dⅹ，令μ=y²→
dμ/dx=(dμ/dy)(dy/dⅹ)=2y(dy/dⅹ)
ⅹ² y²=4→
dⅹ²/dⅹ dy²/dⅹ=d4/dⅹ→
2ⅹ 2ydy/dⅹ=0→
dy/dⅹ=-ⅹ/y
點A(1，√3)為該圓上的點，則該點斜率為dy/dⅹ=-ⅹ/y=-1/√3→
該點切線為(y-√3)/(ⅹ-1)=-1/√3，→y=-(√3/3)x 4√3/3
2y siny=(x²/π) 1→
d(2y)/dⅹ d(siny)/dⅹ=
d(ⅹ²/π)/dⅹ d(1)/dⅹ→
2dy/dⅹ cosydy/dⅹ=2ⅹ/π→
d(2dy/dⅹ)/dⅹ
d(cosydy/dⅹ)/dⅹ=d(2ⅹ/π)/dⅹ
→2d²y/dⅹ² ds/dⅹ=2/π①
s=cosydy/dⅹ
令u=cosy，m=dy/dⅹ→s=um
ds/dⅹ=mdu/dⅹ udm/dⅹ=
(dy/dⅹ)(du/dⅹ) cosyd²y/dⅹ²
du/dⅹ=dcosy/dⅹ=
(dcosy/dy)(dy/dx)=-sinydy/dⅹ→
ds/dⅹ=
-siny(dy/dⅹ)² cosyd²y/dⅹ²→①為
d²y/dⅹ²(2 cosy) -siny(dy/dⅹ)²=
π/2(d²y/dⅹ²與(dy/dⅹ)²不同)
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