Question

求一阶导数y=arccosx/√1-x^2

Answer 1

y=[arccos(x/√1-x^2)]

(cosy)^2=x^2/(1-x^2)

1/(cosy)^2=(1-x^2)/x^2

1/(cosy)^2=1/x^2-1

隐函数求导!

[1/(cosy)^2]'=[1/x^2-1]'

[1/(cosy)^2]'=-2/x^3

S=t^(-2)

t=cosy

y=f(x)

复合函数求导

(S't)*(t'y)*(y'x)=-2/x^3

[－2t^(-3)]*(siny)*(y'x)=-2x^3

[－2(cos[arccos(x/√1-x^2)])^(-3)]*(sin[arccos(x/√1-x^2)])*(y'x)=-2x^3

(y'x)=x^3/{[(cos[arccos(x/√1-x^2)])^(-3)]*(sin[arccos(x/√1-x^2)])}