# Question #fbdf4

Oct 24, 2016

$\frac{d}{\mathrm{dx}} 5 \ln \left(5 x\right) = \frac{5}{x}$

#### Explanation:

Depends on what tools you have available.

Using the chain rule, along with the known derivative $\frac{d}{\mathrm{dx}} \ln x = \frac{1}{x}$:

$\frac{d}{\mathrm{dx}} 5 \ln \left(5 x\right) = 5 \frac{d}{\mathrm{dx}} \ln \left(5 x\right)$

$= 5 \left(\frac{1}{5 x}\right) \left(\frac{d}{\mathrm{dx}} 5 x\right)$

$= 5 \left(\frac{1}{5 x}\right) \left(5\right)$

$= \frac{5}{x}$

Using implicit differentiation:

Let $y = 5 \ln \left(5 x\right)$

$\implies {e}^{y} = {e}^{5 \ln \left(5 x\right)} = {e}^{\ln \left({\left(5 x\right)}^{5}\right)} = {\left(5 x\right)}^{5} = {5}^{5} {x}^{5}$

$\implies \frac{d}{\mathrm{dx}} {e}^{y} = \frac{d}{\mathrm{dx}} {5}^{5} {x}^{5}$

$\implies {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} = {5}^{6} {x}^{4}$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{5}^{6} {x}^{4}}{e} ^ y$

$= \frac{{5}^{6} {x}^{4}}{{5}^{5} {x}^{5}}$

$= \frac{5}{x}$

Using the definition of a derivative:

${\lim}_{h \to 0} \frac{5 \ln \left(5 \left(x + h\right)\right) - 5 \ln \left(5 x\right)}{h}$

$= 5 {\lim}_{h \to 0} \frac{\ln \left(5 \left(x + h\right)\right) - \ln \left(5 x\right)}{h}$

$= 5 {\lim}_{h \to 0} \ln \frac{\frac{5 \left(x + h\right)}{5 x}}{h}$

$= 5 {\lim}_{h \to 0} \frac{1}{h} \ln \left(1 + \frac{h}{x}\right)$

$= 5 {\lim}_{h \to 0} \ln \left[{\left(1 + \frac{h}{x}\right)}^{\frac{1}{h}}\right]$

$= 5 {\lim}_{h \to 0} \ln \left[{\left(1 + \frac{\frac{1}{x}}{\frac{1}{h}}\right)}^{\frac{1}{h}}\right]$

Substitute $u = \frac{1}{h}$. Then $u \to \infty$ as $h \to 0$. Recall ${\lim}_{n \to \infty} {\left(1 + \frac{x}{n}\right)}^{n} = {e}^{x}$.

$= 5 {\lim}_{u \to \infty} \ln \left[{\left(1 + \frac{\frac{1}{x}}{u}\right)}^{u}\right]$

$= 5 \ln \left({e}^{\frac{1}{x}}\right)$

$= \frac{5}{x} \ln \left(e\right)$

$= \frac{5}{x}$