If $y=ae^{-\lambda x}+be^{-\mu x}+ce^{-\nu x}$ when does $\frac{dy}{dx}=0$ How can you find the turning point, such that $\frac{dy}{dx}=0$, of
$$y=ae^{-\lambda x}+be^{-\mu x}+ce^{-\nu x}$$
for $x\in\mathbb{R}$ and where $a,b,c,\lambda,\mu,\nu \in \mathbb{R}$ and $\lambda,\mu,\nu > 0$.
 A: HINT:
$$\frac{\partial}{\partial x}\left[\frac{a}{e^{\lambda x}}+\frac{b}{e^{\mu x}}+\frac{c}{e^{\nu x}}\right]=0\Longleftrightarrow$$
$$\frac{\partial}{\partial x}\left[\frac{a}{e^{\lambda x}}\right]+\frac{\partial}{\partial x}\left[\frac{b}{e^{\mu x}}\right]+\frac{\partial}{\partial x}\left[\frac{c}{e^{\nu x}}\right]=0\Longleftrightarrow$$
$$a\cdot\frac{\partial}{\partial x}\left[\frac{1}{e^{\lambda x}}\right]+b\cdot\frac{\partial}{\partial x}\left[\frac{1}{e^{\mu x}}\right]+c\cdot\frac{\partial}{\partial x}\left[\frac{1}{e^{\nu x}}\right]=0\Longleftrightarrow$$
$$a\cdot\left[-\frac{\lambda}{e^{\lambda x}}\right]+b\cdot\left[-\frac{\mu}{e^{\mu x}}\right]+c\cdot\left[-\frac{\nu}{e^{\nu x}}\right]=0\Longleftrightarrow$$
$$-\frac{a\lambda}{e^{\lambda x}}-\frac{b\mu}{e^{\mu x}}-\frac{c\nu}{e^{\nu x}}=0\Longleftrightarrow$$
$$\frac{a\lambda}{e^{\lambda x}}+\frac{b\mu}{e^{\mu x}}+\frac{c\nu}{e^{\nu x}}=0\Longleftrightarrow$$
$$a\lambda e^{x(\mu+\nu)}+b\mu e^{x(\lambda+\nu)}+c\nu e^{x(\lambda+\mu)}=0$$
