implicitly differentiate y=e^(-mx)

Question

I don't get how to do this and I'm not sure how to use the ln 

 

y' = 

Answer 1

Standard developments of calculus, like Apostol and like Spivak, define log in terms of an integral, find its derivative, then use that to find the derivative of e^f(x). That may be the motivation for doing things this way here.

 

In this problem,

 

ln y = ln e^-mx

       = -mx

 

d/dx ln y = d/dx (-mx)

1/y * dy/dx = -m

dy/dx = -my

          = -me^-mx

 

Now check that this answer agrees with what we get using the rule d/dx e^f(x) = f'(x)e^f(x).