Question

y=tanx的x次方+x的sinx次方 求dy/dx

Answer 1

两项分别求，取对数，然后解完加起来

Answer 2

y=(tanx)^x+x^(sinx)

Y1=(tanx)^x
lnY1=ln(tanx)^x=xln(tanx)
d(lnY1)/dx=(d(lnY1)/d(Y1)) * (d(Y1)/dx)=(1/(Y1)) * (d(Y1)/dx)
d(xln(tanx))/dx=ln(tanx)+(x/tanx)(secx)^2
d(Y1)/dx=((tanx)^x)(ln(tanx)+(x/tanx)(secx)^2)

同理
Y2=(x)^(sinx)
lnY2=ln(x)^(sinx)=(sinx)ln(x)
d(lnY2)/dx=(d(lnY2)/d(Y2)) * (d(Y2)/dx)=(1/(Y2)) * (d(Y2)/dx)
d((sinx)ln(x))/dx=(cosx)ln(x)+(sinx)/x
d(Y2)/dx=((x)^(sinx))((cosx)ln(x)+(sinx)/x)

dy/dx=d(Y1)/dx + d(Y2)/dx =((tanx)^x)(ln(tanx)+(x/tanx)(secx)^2)+((x)^(sinx))((cosx)ln(x)+(sinx)/x)