# How do you use the chain rule to differentiate root9(e^(6x))?

Mar 16, 2017

$f \left(x\right) = \frac{2}{3} \sqrt[3]{{e}^{2 x}}$

#### Explanation:

According to chain rule when $f = f \left(g \left(h \left(x\right)\right)\right)$,

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dg}} \times \frac{\mathrm{dg}}{\mathrm{dh}} \times \frac{\mathrm{dh}}{\mathrm{dx}}$

Here $f \left(x\right) = \sqrt[9]{g} \left(x\right)$, $g \left(x\right) = {e}^{h \left(x\right)}$ and $h \left(x\right) = 6 x$

and $\frac{\mathrm{df}}{\mathrm{dg}} = \frac{1}{9} \left(g {\left(x\right)}^{\left(\frac{1}{9} - 1\right)}\right)$

$\frac{\mathrm{dg}}{\mathrm{dx}} = {e}^{h \left(x\right)}$ and $\frac{\mathrm{dh}}{\mathrm{dx}} = 6$

Hence as $f \left(x\right) = \sqrt[9]{{e}^{6 x}}$

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{1}{9} \left(g {\left(x\right)}^{\left(\frac{1}{9} - 1\right)}\right) \times {e}^{h \left(x\right)} \times 6$

= $\frac{6}{9} {\left({e}^{6 x}\right)}^{\left(\frac{1}{9} - 1\right)} {e}^{6 x}$

= $\frac{6}{9} {e}^{6 x \times \left(\frac{1}{9} - 1\right)} {e}^{6 x}$

= $\frac{6}{9} {e}^{6 x \times \left(\frac{1}{9} - 1\right) + 6 x}$

= $\frac{6}{9} {e}^{\frac{6 x}{9}}$

= $\frac{2}{3} \sqrt[3]{{e}^{2 x}}$