Date: 02/12/2001 at 09:08:57 From: Julie Bishop Subject: Derivative of non-constant functions Are there any non-constant functions f and g such that (fg)' = f'g'? Thanks, Julie
Date: 02/12/2001 at 12:14:09 From: Doctor Rob Subject: Re: Derivative of non-constant functions Thanks for writing to Ask Dr. Math, Julie. Try f(x) = g(x) = e^(2*x) More generally, try f(x) = a*e^(b*x) g(x) = c*e^(b*x/[b-1]) where a, b, c are nonzero and b is unequal to 1. There are other solutions, too: (f*g)' = f'*g' f*g' + f'*g = f'*g' g'/g + f'/f = (g'/g)*(f'/f) g'/g = f'/(f'-f) Now the integral of the left-hand side is ln|g|. If you can integrate the right-hand side, you can solve for g. For example, if f(x) = a*(x+1), then f'(x) = a, and ln|g(x)| = INT[-1/x dx] = -ln|x| + c |g(x)| = C/|x| g(x) = C/|x| for any constant C. You can check that this pair satisfy the equation by using the facts that |x| = sign(x)*x |x|' = sign(x) except at x = 0 The first example above is obtained by letting f'/f = g'/g, which forced f'/f = 2, f(x) = a*e^(2*x), for any constant a. The second example was found by setting f'/f = b for constant b. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
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