# Question #d9dbd

Dec 13, 2016

$\frac{d}{\mathrm{dx}} \log \left(x - 2\right) = \frac{1}{x - 2}$

#### Explanation:

This answer supposes $\log \left(x\right)$ refers to the natural (base-$e$) logarithm.*

Using the chain rule and the derivative of $\log \left(x\right)$:

The chain rule, together with the known derivative $\frac{d}{\mathrm{dx}} \log \left(x\right) = \frac{1}{x}$, gives us

$\frac{d}{\mathrm{dx}} \log \left(x - 2\right) = \frac{1}{x - 2} \left(\frac{d}{\mathrm{dx}} \left(x - 2\right)\right)$

$= \frac{1}{x - 2} \left(1\right)$

$= \frac{1}{x - 2}$

Using implicit differentiation and the derivative of ${e}^{x}$:

If we do not know the derivative of $\log \left(x\right)$, but do know the derivative $\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x}$, we can first create and modify an equation and use implicit differentiation:

Let $y = \log \left(x - 2\right)$

$\implies {e}^{y} = {e}^{\log} \left(x - 2\right)$

$\implies {e}^{y} = x - 2$

$\implies \frac{d}{\mathrm{dx}} {e}^{y} = \frac{d}{\mathrm{dx}} \left(x - 2\right)$

$\implies {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{e} ^ y$

$= \frac{1}{e} ^ \left(\log \left(x - 2\right)\right)$

$= \frac{1}{x - 2}$

*If the intended logarithm is base-$10$, rather than the natural logarithm, then a coefficient of $\frac{1}{\ln} \left(10\right)$ should be included in the answer, as converting to the natural logarithm would give ${\log}_{10} \left(x\right) = \ln \frac{x}{\ln} \left(10\right)$, and then the process would proceed as above.