Could you differentiate this function? f(x)=6x^2ln(4x)

Dec 16, 2017

$12 x \cdot \ln \left(4 x\right) + 6 x$

Explanation:

first use product rule:

$f ' \left(x\right) = \ln \left(4 x\right) \cdot \frac{d}{\mathrm{dx}} \left(6 {x}^{2}\right) + 6 {x}^{2} \cdot \frac{d}{\mathrm{dx}} \left(\ln \left(4 x\right)\right)$

$= \ln \left(4 x\right) \cdot 12 x + \frac{6 {x}^{2}}{4 x} \cdot \frac{d}{\mathrm{dx}} \left(4 x\right)$ (power rule, ln(x) derivative, and chain rule)

$= \ln \left(4 x\right) \cdot 12 x + \left(\frac{3}{2} x \cdot 4\right)$
$= 12 x \cdot \ln \left(4 x\right) + 6 x$

Dec 17, 2017

$f ' \left(x\right) = 12 \left(1 + \ln 4 x\right)$

Explanation:

$\text{differentiate using the "color(blue)"product rule}$

$\text{given "f(x)=g(x)h(x)" then}$

$f ' \left(x\right) = g \left(x\right) h ' \left(x\right) + h \left(x\right) g ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{product rule}}$

$g \left(x\right) = 6 {x}^{2} \Rightarrow g ' \left(x\right) = 12 x$

.$h \left(x\right) = \ln 4 x \Rightarrow h ' \left(x\right) = \frac{1}{4 x} \times \frac{d}{\mathrm{dx}} \left(4 x\right) = \frac{1}{4 x} \times 4 = \frac{1}{x}$

$\Rightarrow f ' \left(x\right) = 6 {x}^{2} \times \frac{1}{x} + \ln 4 x .12 x$

$= 6 x + 12 x \ln 4 x$

$= 6 x \left(1 + 2 \ln 4 x\right)$