How do you differentiate y=ln(3xe^(1-x))?

Jul 31, 2018

$\frac{1 - x}{x}$.

Explanation:

$y = \ln \left(3 x {e}^{1 - x}\right)$.

Using the usual rules of log function, we have,

$y = \ln 3 + \ln x + \ln {e}^{1 - x} , \mathmr{and} ,$

$y = \ln 3 + \ln x + \left(1 - x\right) \ln e = \ln 3 + \ln x + 1 - x$.

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = 0 + \frac{1}{x} + 0 - 1$.

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - x}{x}$, as Respected Sonnhard has derived!