Question

求微分y=(1+x^2)^tanx,求dy/dx

Answer 1

两边取对数

lny=tanx*ln(1+x^2)

y'*1/y=sec²xln(1+x^2)+tanx *2x/(1+x^2)

y'=y[sec²xln(1+x^2)+2xtanx/(1+x^2)]

∴dy/dx=y[sec²xln(1+x^2)+2xtanx/(1+x^2)]

=(1+x^2)^tan[ sec²xln(1+x^2)+2xtanx/(1+x^2)]