More on derivative of an exponential

[Loren Booda]

We know that d/dx(exp(cx))=c(exp(cx)), where c is a constant.

Does there exist a function g(cx) such that d/dx(exp(g(cx)))=1/c(exp(g(cx))) ?

 

[StephenPrivitera]

If cx=h, then g(h)=h/c².
d[e^g(h)]/dx=(dg/dh)(dh/dx)e^g(h)=(1/c²)(c)e^g(h)=(1/c)e^g(h)

 

[mathman]

If cx=h, then g(h)=h/c².
de^g(h)/dx=(dg/dh)(dh/dx)e^g(h)=(1/c2)(c)e^g(h)=(1/c)e^g(h)

That doesn't really answer the question. All you have is
d(e^x/c/dx=(1/c)e^x/c. In other words, you have changed from c to 1/c.

 

[StephenPrivitera]

I know it's only a small change but it answers the question. Loren wanted a composite function g(h(x)) such that d[e^g]/dx=(1/c)e^g. Clearly dg/dx=1/c. I did forget a constant: g(h(x))=x/c+C, h(x)=cx, g(cx)=x/c+C=cx/c²+C=h/c²+C. Perhaps I misunderstand the question.

 

[Loren Booda]

Perhaps I am asking how one might take a "function of linear constant" in an exponential argument and obtain the reciprocal constant upon differentiation. (I explained the problem better mathematically in my original post.) I tend to agree with mathman here, that I wanted to preserve the linear constant relationship in the argument.

Any solutions?

 

[StephenPrivitera]

The only solution is g(h)=h/c²+C or g(cx)=x/c+C. If h=cx, then you get the desired derivative but you can't preserve the linear relationship (if by that you mean the "cx" term) because the c's cancel. Although, technically, it is linear. Since c is a constant, so is 1/c. But as far as your question goes, there are no solutions that provide what you're looking for.
d[e^g(cx)]/dx=[dg/d(cx)][d(cx)/dx]e^g(cx)=(1/c)e^g(cx)
since e^g(cx) is never zero, you can divide the last two parts of the equation by this term and you get,
[dg/d(cx)][d(cx)/dx]=1/c
[dg/d(cx)]c=1/c
dg/d(cx)=1/c²
dg=[1/c²]d(cx)
let u=cx
du=d(cx)
dg=[1/c²]du
g=u/c²+C=x/c+C
This is the only family of functions that will satisfy that equation.