Question

y=ln(x+√1+X^2)的导数

Answer 1

[x+√(1+x²)]'=x'+[√(1+x²)]'=1+[√(1+x²)]'

关键是后面的[√(1+x²)]'如何计算,用链式法则

令y=√(1+x²), u=1+x², 则

y=√u

∴y'=dy/dx

=(dy/du)*(du/dx)

=[d(√u)/du]*[d(1+x²)/dx]

=[1/(2√u)]*(2x)

=2x/2√u

=2x/2√(1+x²)

=x/√(1+x²)

∴[x+√(1+x²)]'=x'+[√(1+x²)]'=1+[√(1+x²)]'=1+x/√(1+x²)