Differentiate : y = ln (e^(-x) + xe^(-x)

Question

Differentiate :

 

y = ln (e^-x + xe^-x  )

p 223, #19

Answer 1

Hi Mia:

 

when differentiating a logarithm function, say y = ln (u),  then the derivative dy/dx is the ratio of du/dx (the derivative of function u inside the parentheses) and of u(x).

d{ ln(u) }/dx = (du/dx) / u;

 

when differentiating an exponential function, say y = exp(u), the derivative is:

d{ exp(u) }/dx = y * ( du/dx ) = { exp(u) } * (du/dx);

 

when differentiating the product of 2 functions, say y = u * w, then the derivative is:

d{ u * w }/dx = u * (dw/dx) + w * (du/dx);

 

this should help you now solve this problem.  Try it on your own, and let me know if this helps.

 

  Jean-Michel