Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In general, $\frac{d}{dx}(f(x) \cdot g(x)) \neq \frac{d}{dx}f(x) \cdot \frac{d}{dx}g(x)$

When does this result hold true? My first try is to use product rule on left side and compare the two sides, but this hasn't helped at all. Any suggestions?

share|improve this question
If $f(x)=g(x)=e^{2x}$, this is true. –  Your Ad Here Feb 20 '14 at 20:57
is that an exhaustive list? –  jj77646_piyy Feb 20 '14 at 20:58
See the answers here for other examples. –  David Mitra Feb 20 '14 at 21:00

2 Answers 2

well assuming one of the two functions, e.g. $g$ is given, then finding $f$ is just solving a homogeneous linear equation.

$$f'(x)g(x) + f(x)g'(x) = f'(x)g'(x)$$ which is the same as $$f'(x)(g(x)-g'(x)) + f(x)g'(x) = 0$$ given $g$ then you can solve it by $$f(x) = f(x_0)\exp \left\{ -\int_{x_0}^x \frac{g'(y)}{g(y)-g'(y)}dy\right\}.$$

share|improve this answer

One can try the exponential of the form $f(x) = e^{ax}$ and $g(x)= e^{bx}$, then

$(fg)' = f'g+fg' = e^{ax}e^{bx}(a+b)=^! f'g'= ae^{ax}be^{bx}\Leftrightarrow a+b = ab\Leftrightarrow a,b = 0 \lor a,b = 2. $

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.