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

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- Thread starter Loren Booda
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- #1

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

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d[e

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mathman

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That doesn't really answer the question. All you have isIf cx=h, then g(h)=h/c^{2}.

de^{g(h)}/dx=(dg/dh)(dh/dx)e^{g(h)}=(1/c2)(c)e^{g(h)}=(1/c)e^{g(h)}

d(e

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Any solutions?

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The only solution is g(h)=h/c^{2}+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/c^{2}

dg=[1/c^{2}]d(cx)

let u=cx

du=d(cx)

dg=[1/c^{2}]du

g=u/c^{2}+C=x/c+C

This is the only family of functions that will satisfy that equation.

d[e

since e

[dg/d(cx)][d(cx)/dx]=1/c

[dg/d(cx)]c=1/c

dg/d(cx)=1/c

dg=[1/c

let u=cx

du=d(cx)

dg=[1/c

g=u/c

This is the only family of functions that will satisfy that equation.

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