# Question #056b8

Nov 8, 2015

Apply the chain rule twice to find $f ' \left(x\right) = \frac{20 {\ln}^{3} \left(5 x - 1\right)}{5 x - 1}$

#### Explanation:

To solve this, we will use the chain rule, which states
$\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = f ' \left(g \left(x\right)\right) g ' \left(x\right)$
Let
${f}_{1} \left(x\right) = {x}^{4}$
${f}_{2} \left(x\right) = \ln \left(x\right)$
${f}_{3} \left(x\right) = 5 x - 1$
${f}_{4} \left(x\right) = \ln \left(5 x - 1\right) = {f}_{2} \left({f}_{3} \left(x\right)\right)$

Now, applying the chain rule,
$\frac{d}{\mathrm{dx}} {\ln}^{4} \left(5 x - 1\right) = \frac{d}{\mathrm{dx}} {f}_{1} \left({f}_{4} \left(x\right)\right) = {f}_{1} ' \left({f}_{4} \left(x\right)\right) {f}_{4} ' \left(x\right)$

${f}_{1} ' \left(x\right) = 4 {x}^{3}$ so ${f}_{1} ' \left({f}_{4} \left(x\right)\right) = 4 {\ln}^{3} \left(5 x - 1\right)$

To find ${f}_{4} ' \left(x\right)$ we once again use the chain rule.

${f}_{4} ' \left(x\right) = \frac{d}{\mathrm{dx}} {f}_{2} \left({f}_{3} \left(x\right)\right) = {f}_{2} ' \left({f}_{3} \left(x\right)\right) {f}_{3} ' \left(x\right)$

${f}_{2} ' \left(x\right) = \frac{1}{x}$ and ${f}_{3} ' \left(x\right) = 5$

So we have
${f}_{4} ' \left(x\right) = \frac{1}{5 x - 1} \cdot 5 = \frac{5}{5 x - 1}$

Substituting back into the original equation gives us the final result

$\frac{d}{\mathrm{dx}} {\ln}^{4} \left(5 x - 1\right) = 4 {\ln}^{3} \left(5 x - 1\right) \cdot \frac{5}{5 x - 1} = \frac{20 {\ln}^{3} \left(5 x - 1\right)}{5 x - 1}$