f(x) = x^5+3x^3+x → df(x)/dx = 5x^4+9x^2+1
dg(x)/dx = 1 / [ df(x)/dx ] = 1 / (5x^4+9x^2+1)
由 f(x)=5 得 x=1 ,所以
g′(5) = [ 1 / (5*x^4+9*x^2+1) ] ┃x=1
= 1/15
f(x) = x^5+3x^3+x → df(x)/dx = 5x^4+9x^2+1
dg(x)/dx = 1 / [ df(x)/dx ] = 1 / (5x^4+9x^2+1)
由 f(x)=5 得 x=1 ,所以
g′(5) = [ 1 / (5*x^4+9*x^2+1) ] ┃x=1
= 1/15