if you have g(x)=ln(x^10)+e^ln(x)

Question

would you just get 1/x^10+e^ln(x) or am i wrong the way it was explained to me that with e the derivative is just wat it is you don't change it

Answer 1

you missed it.

y=ln (u)
dy/du=1/u
dy/dx=1/u * du/dx

y=ln(x^10)
dy/dx=1/x^10 * 10x^9

now,
z=e^u
z'=e^u
dz/dx=e^u * du/dx

y=e^lnx, then
dy/dx=e^lnx * 1/x= x/x=1