# What is the derivative of f(x) = (x)(6^(-2x))?

## What is the derivative of $f \left(x\right) = \left(x\right) \left({6}^{- 2 x}\right)$?

$f ' \left(x\right) = {6}^{- 2 x} \left(1 - 2 x \setminus \ln \left(6\right)\right)$

#### Explanation:

Using product rule of differentiation

$f \left(x\right) = x {6}^{- 2 x}$

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

$\setminus \frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = x \setminus \frac{d}{\mathrm{dx}} {6}^{- 2 x} + {6}^{- 2 x} \setminus \frac{d}{\mathrm{dx}} x$

$= x {6}^{- 2 x} \setminus \ln \left(6\right) \setminus \cdot \left(- 2\right) + {6}^{- 2 x} \setminus \cdot 1$

$= {6}^{- 2 x} \left(- 2 x \setminus \ln \left(6\right) + 1\right)$

$= {6}^{- 2 x} \left(1 - 2 x \setminus \ln \left(6\right)\right)$

Jun 28, 2018

$f ' \left(x\right) = {6}^{- 2 x} + x \ln \left(\frac{1}{36}\right) {6}^{- 2 x}$

#### Explanation:

Given: $f \left(x\right) = \left(x\right) \left({6}^{- 2 x}\right)$

Use the product rule :

$\frac{d \left(u \cdot v\right)}{\mathrm{dx}} = \frac{\mathrm{du}}{\mathrm{dx}} v + u \frac{\mathrm{dv}}{\mathrm{dx}}$

where, $u = x$ and $v = {6}^{- 2 x}$

$\frac{\mathrm{du}}{\mathrm{dx}} = 1$

For $\frac{\mathrm{dv}}{\mathrm{dx}}$, we must use logarithmic differentiation. Start with:

$v = {6}^{- 2 x}$

take the natural logarithm of both sides:

$\ln \left(v\right) = \ln \left({6}^{- 2 x}\right)$

Use the property of logarithms $\ln \left({a}^{c}\right) = c \ln \left(a\right)$:

$\ln \left(v\right) = \left(- 2 x\right) \ln \left(6\right)$

Differentiate both sides:

$\frac{1}{v} \frac{\mathrm{dv}}{\mathrm{dx}} = - 2 \ln \left(6\right)$

$\frac{1}{v} \frac{\mathrm{dv}}{\mathrm{dx}} = \ln \left(\frac{1}{36}\right)$

Multiply both sides by $v$:

$\frac{\mathrm{dv}}{\mathrm{dx}} = \ln \left(\frac{1}{36}\right) v$

We know that $v = {6}^{- 2 x}$:

$\frac{\mathrm{dv}}{\mathrm{dx}} = \ln \left(\frac{1}{36}\right) {6}^{- 2 x}$

Substituting these back into the product rule:

$\frac{d \left(x \left({6}^{- 2 x}\right)\right)}{\mathrm{dx}} = {6}^{- 2 x} + x \ln \left(\frac{1}{36}\right) {6}^{- 2 x}$

$f ' \left(x\right) = {6}^{- 2 x} + x \ln \left(\frac{1}{36}\right) {6}^{- 2 x}$